general design information.pdf

PCI Handbook/Sixth Edition 11–1

CHAPTER 11 GENERAL DESIGN INFORMATION

11.1 Design Information .......................................................................................................11–2

11.1.1 Dead Weights of Floors, Ceilings, Roofs, and Walls.......................................11–2 11.1.2 Recommended Minimum Uniformly Distributed and Concentrated Live Loads.................................................................................11–3 11.1.3 Beam Design Equations and Diagrams ..........................................................11–5 11.1.4 Camber (Deflection) and Rotation Coefficients for Prestress Force and Loads ...........................................................................11–23 11.1.5 Moments in Beams with Fixed Ends .............................................................11–25 11.1.6 Torsion Diagrams, Reactions, and Rotations................................................11–26 11.1.7 Moving Load Placement for Maximum Moment and Shear ..........................11–27 11.1.8 Moments, Shears, and Deflections in Beams with Overhangs .....................11–28

11.2 Material Properties .....................................................................................................11–29

11.2.1 Table of Concrete Stresses...........................................................................11–29 11.2.2 Concrete Modulus of Elasticity as Affected by Unit Weight and Strength ..............................................................................11–29 11.2.3 Properties and Design Strengths of Prestressing Strand and Wire ..............11–30 11.2.4 Properties and Design Strengths of Prestressing Bars.................................11–31 11.2.5 Typical Stress-Strain Curve, 7-Wire Low-Relaxation Prestressing Strand ...11–32 11.2.6 Transfer and Development Lengths for 7-Wire Uncoated Strand.................11–33 11.2.7 Reinforcing Bar Data .....................................................................................11–34 11.2.8 Location of Reinforcement Confined by Stirrups or Ties...............................11–35 11.2.9 Required Development Lengths for Reinforcing Bars...................................11–36 11.2.10 Common Styles of Structural Welded Wire Reinforcement ..........................11–38 11.2.11 Wire Used in Structural Welded Wire Reinforcement ...................................11–39 11.2.12 Bar Area Equivalents in a One Foot Wide Section .......................................11–40 11.2.13 ACI Required Minimum Reinforcement Areas Per Foot Width of Section ....11–41

11.3 Standard Bolts, Nuts and Washers ............................................................................11–42

11.3.1 Dimensions of Nuts and Bolts .......................................................................11–42 11.3.2 Dimensions of Standard Washers.................................................................11–44

11.4 Welding Information....................................................................................................11–45

11.4.1 Weld Symbols Commonly Used in Precast Construction .............................11–45 11.4.2 Typical Welded Joints in Precast Construction .............................................11–46 11.4.3 Properties of Weld Groups Treated as Lines ................................................11–47

11.5 Section Properties ......................................................................................................11–48

11.5.1 Properties of Geometric Sections..................................................................11–48 11.5.2 Plastic Section Moduli and Shape Factors....................................................11–53

11.6 Metric Conversion.......................................................................................................11–54

11.6.1 Metric Calculations and Example ..................................................................11–54 11.6.2 Conversion from U.S. Customary Units to International System ..................11–55 11.6.3 Preferred SI Units and U.S. Customary Equivalents....................................11–57

First Printing/CD-ROM Edition

11–2 PCI Handbook/Sixth Edition

11.1 DESIGN INFORMATION Design Aid 11.1.1 Dead Weights of Floor, Ceilings, Roofs and Wallsa

Component Load (psf) Component Load

(psf) Component Load (psf)

Ceilings Floor Fill Masonry wallsb Acoustical fiber board 1 Cinder concrete, per inch 9 Clay brick wythes: Gypsum board (per Q/K in. thickness) 0.55 Lightweight concrete, per inch 8 4 in. 39 Mechanical duct allowance 4 Sand, per inch 8 8 in. 79 Plaster on tile or concrete 5 Stone concrete, per inch 12 12 in. 115Plaster on wood lath 8 16 in. 155Suspended steel channel system 2 Floors and floor finishes Suspended mtl lath and cem plaster 15 Asphalt block (2 in.) Q/S in. mortar 30Suspended mtl lath and gyp plaster 10 Cement finish (1 in.), on stone concrete fill 32

Hollow concrete masonry unit wythe: Wythe thickness (in.)

Wood furring suspension system 2.5 Ceramic or quarry tile (E/F in.) on Q/S in. bed 16 Ceramic or quarry tile (E/F in.) on 1 in. bed 23 4 6 8 10 12 Coverings, roof and wall Concrete fill finish (per inch thickness) 12 Hardwood flooring, U/K in. 4Asbestos – cement shingles 4 Linoleum or asphalt tile, Q/F in. 1

Density of unit (105 pcf)

Asphalt shingles 2 Marble and mortar on stone-concrete fill 33 No grout 22 24 31 37 43 Cement tiles 16 48 in. o.c. 29 38 47 55 Slate (per inch thickness) 15 40 in. o.c. 30 40 49 57 Clay tile (for mortar add 10 lb.) Solid flat tile on 1 in. mortar base 23 (grout spacing) Book tile, 2 in. 12 Subflooring, E/F in. 3 32 in. o.c. 32 42 52 61 Book tile, 3 in. 20 Terazzo (1 Q/S in.) directly on slab 19 24 in. o.c. 34 46 57 67 Ludowici 10 Terazzo (1 in.) on stone-concrete fill 32 16 in. o.c. 40 53 66 79 Roman 12 Terazzo (1 in.) 2 in. stone concrete 32 Full grout 55 75 95 115 Spanish 19 Wood block (3 in.) on mastic, no fill 10Composition: Wood block (3 in.) on Q/S in. mortar base 16

Density of unit (125 pcf):

Three-ply ready roofing 1 No grout 26 28 36 44 50 Four-ply felt and gravel 5.5 48 in. o.c. 33 44 54 62 Five-ply felt and gravel 6 Floors, wood joist (no plaster) 40 in. o.c. 34 45 56 65 (grout spacing) Copper on tin 1 Double wood floor 32 in. o.c. 36 47 58 68 Corrugated asbestos-cement roofing 4 24 in. o.c. 39 51 63 75 Deck, metal 20 gage 2.5 16 in. o.c. 44 59 73 87 Deck, metal 18 gage 3 Full grout 59 81 102 123Decking, 2 in. wood (Douglas fir) 5

Joist Size (in.)

12 in. spacing

(psf)

16 in. spacing

(psf)

24 in. spacing

(psf) Decking, 3 in. wood (Douglas fir) 8 2 x 6 6 5 5

Density of unit (135 pcf)

Fiberboard, Q/S in. 0.75 2 x 8 6 6 5 No grout 29 30 39 47 54 Gypsum sheating, Q/S in. 2 2 x 10 7 6 6 48 in. o.c. 36 47 57 66 2 x 12 8 7 6 40 in. o.c. 37 48 59 69 (grout spacing) Insulation, roof boards (per in. thickness) 32 in. o.c. 38 50 62 72 Cellular glass 0.7 Frame partitions 24 in. o.c. 41 54 67 78 Fibrous glass 1.1 Movable steel partitions 4 16 in. o.c. 46 61 76 90 Fiberboard 1.5 Wood or steel studs, Q/S gyp board each side 8 Full grout 62 83 105 127 Perlite 0.8 Wood studs, 2 x 4, unplastered 4 Polystyrene foam 0.2 Wood studs, 2 x 4, plastered one side 12 Urethane foam with skin 0.5 Wood studs, 2 x 4, plastered two sides 20

Solid concrete masonry unit wythe: Wythe thickness (in.)

Plywood (per Q/K in. thickness) 0.4 Rigid insulation, Q/S in. 0.75 Frame walls 4 6 8 10 12 Skylight, metal frame, E/K in. wire glass

8 Exterior stud walls: Density of unit (105 pcf) 32 51 69 87 105

Slate, E/AH in. 7 2 x 4 @ 16 in., T/K in. gyp., insulated, E/K in. siding 11 Slate, Q/F in. 10 2 x 6 @ 16 in., T/K in. gyp., insulated, E/K in. siding 12 Density of unit 38 60 81 102 124Waterproofing membranes (125 pcf) Bituminous, gravel-coated 5.5 Exterior stud walls with brick veneer 48 Bituminous, smooth surface 1.5 Windows, glass, frame, and sash 8 Density of unit 41 64 87 110 133 Liquid applied 1 (135 pcf) Single-ply, sheet 0.7 Wood sheathing (per inch thickness) 3 Wood shingles 3

a. Source: “Minimum Design Loads for Buildings and Other Structures,” ASCE 7-02, American Society of Civil Engineers, Reston, VA. b. Weights of masonry include mortar but not plaster. For plaster, add 5 lb/ft2 for each face plasterEdition, Values given represent averages. In

some cases, there is a considerable range of weight for the same construction.



DESIGN INFORMATION Design Aid 11.1.2 Recommended Minimum Uniformly Distributed and Concentrated Live Loadsa

Occupancy or use Uniform load (psf) Concentrated

load (lb) Apartments (see residential) Access floor systems Computer use 100 2,000 Office use 50 2,000 Armories and drill rooms 150 Assembly areas and theaters Fixed seats (fastened to floor) 60 Lobbies 100 Movable seats 100 Platforms (assembly) 100 Stage floors 150 Balconies (exterior) 100 On one- and two-family residences only, and not exceeding 100 ft2 60 Bowling alleys, poolrooms and similar recreational areas 75 Corridors First floor 100 Other floors, same as occupancy served except as indicated Dance halls and ballrooms 100 Decks (patio and roof) Same as area served, or for the type of occupancy accommodated Dining rooms and restaurants 100 Dwellings (see residential) Elevator machine room grating (on area of 4 in.2) 300 Finish light floor plate construction (on area of 1 in.2) 200 Fire escapes 100 On single-family dwellings only 40 Garages (passenger cars only) 50 Note b Truck and buses Note c Grandstands (see stadium and arena bleachers) Gymnasiums, main floors and balconies (see note e) 100 Handrails, guardrails and grab bars Note i Hospitals Corridors above first floor 80 1,000 Operating room, laboratories 60 1,000 Private rooms 40 1,000 Wards 40 1,000 Hotels (see residential) Libraries Corridors above first floor 80 1,000 Reading rooms 60 1,000 Stack rooms (see note d) 150 1,000 Manufacturing Heavy 250 2,000 Light 125 3,000 Marquees and Canopies 75 Office Buildings Corridors above first floor 80

File and computer rooms shall be designed for heavier loads based on anticipated occupancy 2,000

Lobbies and first floor corridors 100 2,000 Offices 50 2,000 See following page for all notes.



DESIGN INFORMATION Design Aid 11.1.2 Recommended Minimum Uniformly Distributed and Concentrated Live Loadsa (Cont.)

Occupancy or use Uniform load (psf) Concentrated

load (lb) Penal institutions Cell Blocks 40 Corridors 100 Residential Dwellings (one- and two-family) Habitable attics and sleeping areas 30 Uninhabitable attics with storage 20 Uninhabitable attics without storage 10 All other areas except balconies 40 Hotels and multifamily houses Private rooms and corridors serving them 40 Public rooms and corridors serving them 100 Reviewing stands, grandstands and bleachers (see Note e) 100 Roofs Note j Schools Classrooms 40 1,000 Corridors above first floor 80 1,000 First floor corridors 100 1,000 Scuttles, skylight ribs, and accessible ceilings 200 Sidewalks, vehicular driveways, and yards, subject to trucking (see Note f, g) 250 8,000 Stadiums and arenas Bleachers (see Note e) 100 Fixed seats, fastened to floor (see Note e) 60 Stairs and exitways 100 Note h Storage areas above ceilings 20 Heavy 250 Light 125 Storage warehouses (shall be designed for heavier loads if required for

anticipated storage)

Stores Retail First floor 100 1,000 Upper floors 75 1,000 Wholesale, all floors 125 1,000 Vehicle barriers Note i Walkways and elevated platforms (other than exitways) 60 Yards and terraces, pedestrians 100 a. Source: “Minimum Design Loads for Buildings and Other Structures,” ASCE 7-02, American Society of Civil Engineers, Reston, VA. b. Floors in garages or portions of buildings used for the storage of motor vehicles shall be designed for the uniformly distributed live loads

of Design Aid 11.1.2 or the following concen trated load: (1) for passenger cars accommodating not more than nine passengers 3,000 lb. acting on an area of 20 in.2; and (2) mechanical parking structures without slab or deck, passenger car only, 1,500 lb/wheel.

c. Garages accommodating trucks and buses shall be designed in accordance with an approved method which contains provisions for truck and bus loadings.

d. The weight of books and shelving shall be computed using an assumed density of 65 pcf and converted to a uniformly distributed load;this load shall be used if it exceeds 150 pcf.

e. In addition to the vertical live loads, horizontal swaying forces parallel and normal to the length of seats shall be included in the designaccording to the requirements of ANSI/NFPA 102.

f. Other uniform loads in accordance with an approved method which contains provisions for truck loadings shall also be considered whereappropriate.

g. The concentrated wheel load shall be applied on an area of 20 in.2. h. Minimum concentrated load on stair treads on area of 4 in.2 is 300 lb. i. See ASCE 7-02, Section 4.4. j. See ASCE 7-02, Sections 4.3 and 4.9.



DESIGN INFORMATION Design Aid 11.1.3 Beam Design Equations and Diagrams

This visual index assists in quickly locating the desired beam equations. The top row shows the support type, and the left column shows the load. For example, to find the equations for a beam fixed on both ends with a uniform load, go down from Column 4 and right on Row B. Locate 4B on upper right corner (page 11–17).



DESIGN INFORMATION Design Aid 11.1.3 Beam Design Equations and Diagrams (Cont.) 1A

1A.1 SIMPLE BEAM – CONCENTRATED LOAD AT CENTER

R = V ........................................................= P2

Mmax (at point of load)...............................= P4

Mx (when <x2

) .....................................= Px2

∆max (at point of load) ...............................= 3P

48El

∆x (when <x2

)......................................= 2 2Px (3 4x )48El

−

1A.2 SIMPLE BEAM – CONCENTRATED LOAD AT ANY POINT

R1 = V1 (max when a < b).........................= Pb

R2 = V2 (max when a > b).........................= Pa

Mmax (at point of load)...............................= Pab

Mx (when x < a) ........................................= Pbx

∆max a(a 2b)at x when a b

3⎛ ⎞+= >⎜ ⎟⎜ ⎟⎝ ⎠

....... = Pab(a 2b) 3a(a 2b)27El

+ +

∆a (at point of load)...................................= 2 2Pa b

3El

∆x (when x < a).........................................= 2 2 2Pbx ( b x )6El

− −

1A.3 SIMPLE BEAM – TWO EQUAL CONCENTRATED LOADS SYMMETRICALLY PLACED R = V........................................................= P Mmax (between loads) ...............................= Pa Mx (when x < a) ........................................= Px

∆max (when) ..............................................= 2 2Pa (3 4a )24El

−

∆x (when x < a).........................................= 2 2Px (3 a 3a x )6El

− −

∆x [when x > a and < ( − a )]...................= 2 2Pa (3 x 3x a )6El

− −



DESIGN INFORMATION Design Aid 11.1.3 Beam Design Equations and Diagrams (Cont.) 1A,1B

1A.4 SIMPLE BEAM – TWO UNEQUAL CONCENTRATED LOADS UNSYMMETRICALLY PLACED

R1 = V1 .....................................................= 1 2P ( a) P b− +

R2 = V2 .....................................................= 1 2P a P ( b)+ −

Vx [when x > a and < ( b− )] ...................= R1 – P1 M1 (max when R1 < P1) ............................= R1a M2 (max when R2 < P2) ............................= R2b Mx (when x < a) ........................................= R1x Mx [when x > a and < ( b− )]...................= R1x – P1(x – a)

1B.1 SIMPLE BEAM – UNIFORMLY DISTRIBUTED LOAD

R = V ........................................................= w2

Vx .............................................................= w x2

⎛ ⎞−⎜ ⎟⎝ ⎠

Mmax (at center) ........................................= 2w

8

Mx.............................................................= wx ( x)2

−

∆max (at center) .........................................= 45w

384El

∆X .............................................................= 3 2 3wx ( 2 x x )24El

− +

1B.2 SIMPLE BEAM – UNIFORMLY DISTRIBUTED LOAD AND VARIABLE END MOMENTS

R1 = V1 .....................................................= 21M Mw2

−+

R2 = V2 .....................................................= 1 2M Mw2

−−

Vx .............................................................= 1 2M Mw x2

−⎛ ⎞− +⎜ ⎟⎝ ⎠

M3 1 2M Mat x2 w

−⎛ ⎞= +⎜ ⎟⎝ ⎠

...........................= 22

1 2 1 22

M M (M M )w8 2 2w

+ −− +

Mx.............................................................= 211

M Mwx ( x) x M2

−⎛ ⎞− + −⎜ ⎟⎝ ⎠

b (to locate inflection points) ....................= 22

1 2 1 2M M M M4 w w

+ −⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∆x..............................................................= 3 2 31 2 1 1 24M 4M 12M 8M 4Mwx x 2 x x24El w w w w w

⎡ ⎤⎛ ⎞− + − + + − −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦



DESIGN INFORMATION Design Aid 11.1.3 Beam Design Equations and Diagrams (Cont.) 1C,1D

1C.1 SIMPLE BEAM – UNIFORM LOAD PARTIALLY DISTRIBUTED

R1 = V1 (max when a < c).........................= wb (2c b)2

+

R2 = V2 (max when a > c).........................= wb (2a b)2

+

Vx [when x > a and < (a + b)]....................= R1 – w(x – a)

Mmax 1Rat x aw

⎛ ⎞= +⎜ ⎟⎝ ⎠

................................= 11

RR a2w

⎛ ⎞+⎜ ⎟⎝ ⎠

Mx (when x < a) ........................................= R1x

Mx [when x > a and < (a + b)] ...................= 21

wR x (x a)2

− −

Mx [when x > (a + b)] ................................= 2R ( x)−

1D.1 SIMPLE BEAM – LOAD INCREASING UNIFORMLY TO ONE END (W IS TOTAL LOAD)

W..............................................................= w2

R1 = V1 .....................................................= W3

R2 = V2 (max) ...........................................= 2W3

Vx .............................................................= 2

2W Wx3

−

Mmax at x 0.57743

⎛ ⎞= =⎜ ⎟⎝ ⎠

....................= 2W 0.1283W9 3

=

Mx.............................................................= 2 22

Wx ( x )3

−

∆max 8at x 1 0.5193

15

⎛ ⎞⎜ ⎟= − =⎜ ⎟⎝ ⎠

..........= 3W0.01304

El

∆x..............................................................= 4 2 2 42

Wx (3x 10 x 7 )180El

− +

1D.2 SIMPLE BEAM – LOAD INCREASING UNIFORMLY TO CENTER (W IS TOTAL LOAD)

W..............................................................= w2

R = V ........................................................= W2

Vx when x2

⎛ ⎞<⎜ ⎟⎝ ⎠

.....................................= 2 22

W ( 4x )2

−

Mmax (at center) ........................................= W6

Mx when x2

⎛ ⎞<⎜ ⎟⎝ ⎠

.....................................= 2

21 2xWx2 3

⎛ ⎞−⎜ ⎟⎜ ⎟

⎝ ⎠

∆max (at center) .........................................= 3W

60El

∆x when x2

⎛ ⎞<⎜ ⎟⎝ ⎠

......................................= 2 2 2Wx (5 4x )480El

−



DESIGN INFORMATION Design Aid 11.1.3 Beam Design Equations and Diagrams (Cont.) 1E

1E.1 BEAM SIMPLY SUPPORTED AT BOTH ENDS – MOMENT APPLIED AT ONE END

–R1 = R2 = V..............................................................= oM

Mmax (at R1) ................................................................= Mo

Mx = Mo – R1x .............................................................= oxM 1⎛ ⎞−⎜ ⎟

⎝ ⎠

∆max (when x 0.422= )..............................................= 2

oM0.0642El

∆x................................................................................= 3

2oM x3x 2 x6El

⎛ ⎞− −⎜ ⎟⎜ ⎟

⎝ ⎠

θ1 (at R1).....................................................................= oM3El

θ2 (at R2).....................................................................= oM6El

1E.2 BEAM SIMPLY SUPPORTED AT BOTH ENDS – MOMENT APPLIED AT ANY POINT

R1 = V (when a > b)....................................................= oM

R2 (when a > b) ..........................................................= oM

Mmax(–) (at x = a) .........................................................= oM a

Mmax(+) (at x = a) .........................................................= oaM 1⎛ ⎞−⎜ ⎟

⎝ ⎠

Mx (when x < a) ..........................................................= oM x

Mx (when x > a) ..........................................................= oxM 1⎛ ⎞−⎜ ⎟

⎝ ⎠

∆x (when x < a)...........................................................= ( )2 2 2oM x 3b x6El

− −

∆x (when x > a)...........................................................= 2 2oM ( x) (3a 2 x x )6El

−− +

∆max 2 23bat x if a 0.4226

3

⎛ ⎞−⎜ ⎟= >⎜ ⎟⎝ ⎠

......................=

32 2 2

oM 3b3El 3

⎛ ⎞−⎜ ⎟⎜ ⎟⎝ ⎠

∆max 2 23aat x if a 0.5774

3

⎛ ⎞−⎜ ⎟= − >⎜ ⎟⎝ ⎠

.................=

32 2 2

oM 3a3El 3

⎛ ⎞−⎜ ⎟⎜ ⎟⎝ ⎠

CLM (at center) .........................................................= oM2

−

CL∆ (at center) .........................................................= 2 2oM ( 4b )16El

−

∆max (when a = b = 2

, at x = 3 0.288676

= ) ........= 2

oM124.71El

CLθ (at center) ..........................................................= oM12El



DESIGN INFORMATION Design Aid 11.1.3 Beam Design Equations and Diagrams (Cont.) 1E,2A

1E.3 BEAM SIMPLY SUPPORTED AT BOTH ENDS – MOMENTS APPLIED AT EACH END

R1 = –R2 = V................................................ = 2 1M M−

Mx................................................................ = 2 1 1x(M M ) M− +

∆x..................................................................= 1 2x( x) [M (2 x) M ( x)]

6El− − + +

at x1 = 2 2 2

1 1 1 2 1 2

1 2

6M 36M 12(M M ) (2M M )6(M M )

± − − +−

,

∆ = max and θ = 0

θ1 (at end) ................................................... = 1 2(2M M )6El

− +

θ2 (at end) ................................................... = 1 2(M 2M )6El

+

If M1 and M2 are of opposite signs, the above formulas hold; just use actual sign of moment. Mx (at point of contraflexure),

1

2 1

Mwhere xM M

⎛ ⎞=⎜ ⎟−⎝ ⎠

.................................. = 0

2A.1 BEAM OVERHANGING ONE SUPPORT – CONCENTRATED LOAD AT ANY POINT BETWEEN

SUPPORTS

R1 = V1 (max when a < b).............................= Pb

R2 = V2 (max when a > b).............................= Pa

Mmax (at point of load)...................................= Pab

Mx (when x < a) ............................................= Pbx

∆max a(a 2b)AT x when a b

3⎛ ⎞+= >⎜ ⎟⎜ ⎟⎝ ⎠

.........= Pab(a 2b) 3a(a 2b)27El

+ +

∆a (at point of load).......................................= 2 2Pa b

3El

∆x (when x < a).............................................= 2 2 2Pbx ( b x )6El

− −

∆x (when x > a).............................................= 2 2Pa( x) (2 x x a )6El

− − −

1x∆ ...............................................................= 1Pabx ( a)6El

+




2A.2 BEAM OVERHANGING ONE SUPPORT – CONCENTRATED LOAD AT END OF OVERHANG

R1 = V1 ........................................................ = Pa

R2 = V1 + V2................................................. = P ( a)+

V2 ................................................................ = P Mmax (at R2) ................................................. = Pa

Mx (between supports) ................................ = Pax

1xM (for overhang)...................................... = P(a – x1)

∆max between supports x3

⎛ ⎞=⎜ ⎟⎝ ⎠

............... = =2 2Pa Pa0.06415

El9 3El

∆max (for overhang at x1 = a)........................ = 2Pa ( a)

3El+

∆x (between supports) ................................. = 2 2Pax ( x )6El

−

1x∆ (for overhang) ....................................... = 211 1

Px (2a 3ax x )6El

+ −

2B.1 BEAM OVERHANGING ONE SUPPORT – UNIFORMLY DISTRIBUTED LOAD

R1 = V1 ........................................................ = 2 2w ( a )2

−

R2 = V2 + V3................................................. = 2w ( a)2

+

V2 ................................................................ = wa

V3 ................................................................ = 2 2w ( a )2

+

Vx (between supports)................................. = R1 – wx

1xV (for overhang) ...................................... = w(a – x1)

M1 2

2aat x 1

2

⎛ ⎞⎡ ⎤⎜ ⎟= −⎢ ⎥⎜ ⎟⎢ ⎥⎣ ⎦⎝ ⎠

................................ = 2 22

w ( a) ( a)8

+ −

M2 (at R2) .................................................... = 2wa

2

Mx (between supports) ................................ = 2 2wx ( a x )2

− −

1xM (for overhang)...................................... = 21

w (a x )2

−

∆x (between supports) ................................. = 4 2 2 3 2 2 2 2wx ( 2 x x 2a 2a x )24El

− + − +

1x∆ (for overhang) ...................................... = 2 3 2 2 311 1 1

wx (4a 6a x 4ax x )24El

− + − +



DESIGN INFORMATION Design Aid 11.1.3 Beam Design Equations and Diagrams (Cont.) 2C

2C.1 BEAM OVERHANGING ONE SUPPORT – UNIFORMLY DISTRIBUTED LOAD BETWEEN SUPPORTS

R = V .............................................................. = w2

Vx ................................................................... = w x2

⎛ ⎞−⎜ ⎟⎝ ⎠

Mmax (at center) .............................................. = 2w

8

Mx................................................................... = wx ( x)2

−

∆max (at center) ............................................... = 45w

384El

∆x.................................................................... = 3 2 3wx ( 2 x x )24El

− +

1x∆ ................................................................. = 3

1w x24El

2C.2 BEAM OVERHANGING ONE SUPPORT – UNIFORMLY DISTRIBUTED LOAD ON OVERHANG

R1 = V1 ........................................................... = 2wa

2

R2 = V1 + V2.................................................... = wa (2 a)2

+

V2 ................................................................... = wa

1xV (for overhang) ......................................... = w(a – x1)

Mmax (at R2) .................................................... = 2wa

2

Mx (between supports) ................................... = 2wa x

2

1xM (for overhang)......................................... = 21

w (a x )2

−

∆max (between supports at x = 3

) ............... = 2 2 2 2wa wa0.03208

El18 3El=

∆max (for overhang at x1 = a)........................... = 3wa (4 3a)

24El+

∆x (between supports) .................................... = 2

2 2wa x ( x )12El

−

1x∆ (for overhang) ......................................... = 2 2 2 311 1 1

wx (4a 6a x 4ax x )24El

+ − +




3A.1 BEAM FIXED AT ONE END, SIMPLY SUPPORTED AT THE OTHER END – CONCENTRATED LOAD AT CENTER

R1 = V1 ........................................................... = 5P16

R2 = V2 (max) ................................................. = 11P16

Mmax (at fixed end).......................................... = 3P16

M1 (at point of load)........................................ = 5P32

Mx (when x < 2

)............................................ = 5Px16

Mx (when x > 2

)............................................ = 11xP2 16

⎛ ⎞−⎜ ⎟⎝ ⎠

∆max (at x = 1 0.44725

= ) .......................... = 3 3P P0.009317

El48El 5=

∆x (at point of load)......................................... = 37P

768El

∆x (when x <2

) ............................................. = ( )2 2xP 3 5x96El

−

∆x (when x > 2

) ............................................ = ( ) ( )2P x 11x 296El

− −

3A.2 BEAM FIXED AT ONE END, SIMPLY SUPPORTED AT THE OTHER END – CONCENTRATED LOAD

AT ANY POINT

R1 = V1 ........................................................... = ( )2

3Pb a 22

+

R2 = V2 ........................................................... = ( )2 23

Pa 3 a2

−

M1 (at point of load)........................................ = 1R a

M2 (at fixed end)............................................. = ( )2Pab a2

+

Mx (when x < a) .............................................. = 1R x

Mx (when x > a) ............................................. = ( )1R x P x a− −

∆max (when a < 0.414 , AT x = 2 2

2 2a

3 a+−

)..... = ( )( )

32 2

22 2

Pa a

3El 3 a

−

−

∆max (when a > 0.414 , AT x = a2 a+

) ...... = 2Pab a

6EI 2 a+

∆a (at point of load)......................................... = ( )2 3

3Pa b 3 a12EI

+

∆x (when x < a)............................................... = ( )2

2 2 23

Pb x 3a 2 x ax12EI

− −

∆x (when x > a)............................................... = 2 2 2 22

Pa ( x) (3 x a x 2a )12EI

− − −



DESIGN INFORMATION Design Aid 11.1.3 Beam Design Equations and Diagrams (Cont.) 3B,3C

3B.1 BEAM FIXED AT ONE END, SIMPLY SUPPORTED AT THE OTHER END – UNIFORMLY DISTRIBUTED LOAD

R1 = V1 ......................................................= 3w8

R2 = V2 (max) ............................................= 5w8

Vx ..............................................................= R1 – wx

Mmax ..........................................................= 2w

8

M1 (at x = 38

) ..........................................= 29 w128

Mx..............................................................= 2

1wxR x

2−

∆max (at x = + =(1 33 ) 0.4215 )16

.........= 4w

185El

∆x...............................................................= 3 2 3wx ( 3 x 2x )48El

− +

3C.1 BEAM FIXED AT ONE END, SIMPLY SUPPORTED AT THE OTHER END – UNIFORM LOAD

PARTIALLY DISTRIBUTED OVER SPAN

R1 = V1 ......................................................= 2 3 23

wb (12e 4e b d)8

− +

R2 = V2 ......................................................= wb – R1

Mmax(–)........................................................= 2 3 2 22

wb (12e 4e b d 8e )8

− + −

M1..............................................................= 11

RR a2w

⎛ ⎞+⎜ ⎟⎝ ⎠

Mx (when x < a) .........................................= R1x

Mx [when x > a and x < (a + b)] .................= 21

wR x (x a)2

− −

Mx (when x > (a + b) and x < )...............= R1x – wb(x – d)

∆x (when x < a)..........................................= 2 2 2 21

x [4R (x 3 ) wb(b 12e )]24El

− + +

∆x [when x > a and x < (a + b)] ..................= 2 2 2 2 41

1 [4R x(x 3 ) wbx(b 12e ) w(x a) ]24El

− + + − −

∆x (when x > (a + b) and x < c).................= 2 3MAX 2

1 [3M ( x) R ( x) ]6El

− + −



DESIGN INFORMATION Design Aid 11.1.3 Beam Design Equations and Diagrams (Cont.) 3E,3F

3E.1 BEAM FIXED AT ONE END, SIMPLY SUPPORTED AT THE OTHER END – MOMENT APPLIED AT THE FLEXIBLE END

R1 = –R2 = V..............................................= o3M2

M1..............................................................= Mo M2..............................................................= Q/SMo

Mx..............................................................= oM (2 3x)2

−

∆max (at x = 3

)..........................................= 2

oM27El

∆x...............................................................= 2oM x ( x)4El

−

θ (at supported end)..................................= oM4El

3F.1 BEAM FIXED AT ONE END – DIFFERENTIAL SETTLEMENT OF SUPPORTS

V = R1 = R2................................................= 2 133El ( )∆ − ∆

Mmax ..........................................................= 2 123El ( )∆ − ∆

Mx..............................................................= maxxM 1⎛ ⎞−⎜ ⎟

⎝ ⎠

∆x...............................................................= 2 3

2 11

x x32

⎡ ⎤∆ − ∆ ⎛ ⎞ ⎛ ⎞⎢ ⎥∆ + −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

3F.2 BEAM FIXED AT ONE END – ROTATION OF SUPPORT

V = –R1 = R2..............................................= 123El φ

Mmax ..........................................................= 13El φ

Mx..............................................................= maxxM 1⎛ ⎞−⎜ ⎟

⎝ ⎠

∆max ...........................................................= 1 5.196⎡ ⎤φ ⎢ ⎥⎣ ⎦

∆x...............................................................= 2 3

1 23x xx2 2

⎡ ⎤φ − + −⎢ ⎥⎢ ⎥⎣ ⎦




4A.1 BEAM FIXED AT BOTH ENDS – CONCENTRATED LOAD AT CENTER total equivalent uniform load ....................... = P

R = V ........................................................... = P2

Mmax (at center and ends) ........................... = P8

Mx when x2

⎛ ⎞<⎜ ⎟⎝ ⎠

....................................... = P (4x )8

−

∆max (at center) ............................................ = 3P

192El

∆x when x2

⎛ ⎞<⎜ ⎟⎝ ⎠

......................................... = 2Px (3 4x)

48El−

4A.2 BEAM FIXED AT BOTH ENDS – CONCENTRATED LOAD AT ANY POINT

R1 = V1 (max when a < b)............................ = 2

3Pb (3a b)+

R2 = V2 (max when a > b)............................ = 2

3Pa (a 3b)+

M1 (max when a < b) ................................... = 2

2Pab

M2 (max when a > b) ................................... = 2

2Pa b

Ma (at point of load)..................................... = 2 2

32Pa b

Mx (when x < a) ........................................... = 2

1 2PabR x −

∆max (when a > b, at x = 2a3a b+

) .................. = 3 2

22Pa b

3El(3a b)+

∆a (at point of load)...................................... = 3 3

3Pa b3El

∆x (when x < a)............................................ = 2 2

3Pb x (3a 3ax bx)6El

− −



DESIGN INFORMATION Design Aid 11.1.3 Beam Design Equations and Diagrams (Cont.) 4B,4C

4B.1 BEAM FIXED AT BOTH ENDS – UNIFORMLY DISTRIBUTED LOADS

R = V ....................................................... = w2

Vx ............................................................ = w x2

⎛ ⎞−⎜ ⎟⎝ ⎠

Mmax (at ends) ......................................... = 2w

12

M1 (at center) .......................................... = 2w

24

Mx............................................................ = 2 2w (6 x 6x )12

− −

∆max (at center) ........................................ = 4w

384El

∆x............................................................. = 2

2wx ( x)24El

−

4C.1 BEAM FIXED AT BOTH ENDS – UNIFORM LOAD PARTIALLY DISTRIBUTED OVER SPAN

R1 = V1 .................................................... = 2 23

wb [4e ( 2d) b (c a)]4

+ − −

R2 = V2 .................................................... = wb – R1

M1............................................................ = 2 22

wb {b [ 3(c a)] 24e d}24

+ − −

M2............................................................ = 1R – wbe + M1

Mmax(+) 1Rat x aw

⎛ ⎞= +⎜ ⎟⎝ ⎠

........................... = 11 1

RM R a2w

⎛ ⎞+ +⎜ ⎟⎝ ⎠

Mx (when x < a) ....................................... = M1 + R1x

Mx [when x > a and x < (a + b)] ............... = 21 1

wM R x (x a)2

+ − −

∆x (when x < a)........................................ = 2 31 1

1 (3M x R x )6El

+

∆x [when x > a and x < (a + b)] ................ = 2 3 41 1

1 [12M x 4R x w(x a) ]24El

+ − −



DESIGN INFORMATION Design Aid 11.1.3 Beam Design Equations and Diagrams (Cont.) 4E,4F

4E.1 BEAM FIXED AT BOTH ENDS – MOMENT APPLIED AT ANY POINT

R1 = V................................................= o3

6M ab−

R2 ......................................................= o3

6M ab

M1......................................................= o2

M b ( 3a)− −

M2......................................................= o2

M a (2 3a)− −

Mx (when x < a) .................................= o2

M 6abx b( 3a)⎡ ⎤− + −⎢ ⎥⎣ ⎦

Mx (when x > a) .................................= o2

M a 6bx6b 2 3a⎛ ⎞− − +⎜ ⎟⎝ ⎠

Mmax(–) (at x = a on left side) ..............= Mmax(+) – Mo

Mmax(+) (at x = a on right side) ............= 2

o 3 26a b bM ( 3a) 1

⎡ ⎤− − − +⎢ ⎥⎢ ⎥⎣ ⎦

∆x (when x < a)..................................= 2

o2

M bx 2ax3a2El

⎛ ⎞− − +⎜ ⎟⎝ ⎠

∆x (when x > a)..................................= 2

o2

M a( x) 2bx3a 2 2b2El

− ⎛ ⎞− + −⎜ ⎟⎝ ⎠

CLM (at center).................................= o2

M [3ab b( 3a)]− + −

CL∆ (at center) .................................= oM b ( 2a)8El

− −

∆max (when a = 0.2324l ) .................= 2

o0.01615MEl

−

4F.1 BEAM FIXED AT BOTH ENDS – DIFFERENTIAL SETTLEMENT OF SUPPORTS

V = –R1 = R2......................................= 2 1312El ( )∆ − ∆

M1 = –M2 ...........................................= 2 126El ( )∆ − ∆

Mx......................................................= 2 126El 2x( ) 1⎛ ⎞∆ − ∆ −⎜ ⎟

⎝ ⎠

∆x.......................................................= 2 3

1 2 1x x( ) 3 2

⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥∆ + ∆ − ∆ −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦



DESIGN INFORMATION Design Aid 11.1.3 Beam Design Equations and Diagrams (Cont.) 4F,5A

4F.2 BEAM FIXED AT BOTH ENDS – ROTATION OF SUPPORT

V = –R1 = R2..............................................= 226El φ

M1..............................................................= 22El φ

M2..............................................................= 24El φ

Mx..............................................................= 22El 3x1⎛ ⎞φ −⎜ ⎟

⎝ ⎠

∆max (at x = 23

)........................................= 24

27− φ

∆x...............................................................= 2 3

2x x⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥− φ −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

5A.1 CANTILEVER BEAM – CONCENTRATED LOAD AT FREE END R = V .........................................................= P Mmax (at fixed end).....................................= P Mx..............................................................= Px

∆max (at free end).......................................= 3P

3El

∆x...............................................................= 3 2 3P (2 3 x x )6El

− +

5A.2 CANTILEVER BEAM – CONCENTRATED LOAD AT ANY POINT R = V .........................................................= P Mmax (at fixed end).....................................= Pb Mx (when x > a) .........................................= P(x – a)

∆max (at free end).......................................= 2Pb (3 b)

6El−

∆a (at point of load)....................................= 3Pb

3El

∆x (when x < a)..........................................= 2Pb (3 3x b)

6El− −

∆x (when x > a)..........................................= 2P( x) (3b x)

6El− − +



DESIGN INFORMATION Design Aid 11.1.3 Beam Design Equations and Diagrams (Cont.) 5B,5C,5D

5B.1 CANTILEVER BEAM – UNIFORMLY DISTRIBUTED LOAD R = V ..........................................= w Vx ...............................................= wx

Mmax (at fixed end)......................= 2w

2

Mx...............................................= 2wx

2

∆max (at free end)........................= 4w

8El

∆x................................................= 4 3 4w (x 4 x 3 )24El

− +

5C.1 CANTILEVER BEAM – UNIFORM LOAD PARTIALLY DISTRIBUTED AT FREE END R = V ..........................................= wb Mmax (at support) ........................= wbe

Mx (when x < b) ..........................= 2wx

2

Mx (when x > b) ..........................= wb (b 2x)2

−

∆max (at free end)........................= 3 2 3wb (8e 24e b )48El

− −

∆x (when x < b)...........................= 3 2 3 4 4w [8be 24be ( x) 2b x b 2x ]48El

− − + − −

∆x (when x > b)...........................= 3 2 3wb [8e 24e ( x) (2x b) ]48El

− − − −

θ (at free end).............................= 2 2wb (b 12e )24El

+

5D.1 CANTILEVER BEAM – LOAD INCREASING UNIFORMLY TO FIXED END (W IS TOTAL LOAD)

W................................................= w2

R = V ..........................................= W

Vx ...............................................= 2

2xW

Mmax (at fixed end)......................= W3

Mx...............................................= 3

2Wx3

∆max (at free end)........................=3W

15El

∆x................................................= 5 4 52

W (x 5 x 4 )60El

− +



DESIGN INFORMATION Design Aid 11.1.3 Beam Design Equations and Diagrams (Cont.) 5D,5E

5D.2 CANTILEVER BEAM – VARYING LOAD INCREASING UNIFORMLY FROM SUPPORT TO FREE END (W IS TOTAL LOAD)

W.....................................................= w2

R = V ...............................................= W

Vx ....................................................= 22Wx x

2⎛ ⎞−⎜ ⎟⎝ ⎠

Mmax (at support) ............................. = 2W3

Mx....................................................= 2

2Wx (x 3 )3

−

∆max (at free end)............................. = 311W

60El

∆x.....................................................= 4 42

W [ (15x 11 ) x (5 x)]60El

− − −

θ (at free end).................................. = 2W

4El

5E.1 CANTILEVER BEAM – MOMENT APPLIED AT FREE END R = V ...............................................= O Mx....................................................= Mo

∆max (at free end)............................. = 2

oM2El

∆x.....................................................= 2oM ( x)2El

−

θ (at free end).................................. = oMEl




6A.1 TWO SPANS, CONTINUOUS BEAM – CONCENTRATED LOAD AT CENTER OF ONE SPAN ONLY

R1 = V1 ........................................................ = 13 P32

R2 =V2 + V3.................................................. = 11P16

R3 = V3 ........................................................ = 3 P32

−

V2 ................................................................ = 19 P32

Mmax (at point of load).................................. = 13 P64

M2 (at R2) .................................................... = 3 P32

6A.2 TWO SPANS, CONTINUOUS BEAM – CONCENTRATED LOAD AT ANY POINT OF ONE SPAN ONLY

R1 = V1 ........................................................ = 23

Pb [4 a( a)]4

− +

R2 = V2 + V3................................................. = 23

Pa [2 b( a)]2

+ +

R3 = V3 ........................................................ = 3Pab ( a)4

− +

V2 ................................................................ = 23

Pa [4 b( a)]4

⎛ ⎞− − +⎜ ⎟⎝ ⎠

Mmax (at point of load).................................. = 23

Pab [4 a( a)]4

− +

M2 (at R2) .................................................... = 2Pab ( a)4

+

6B.1 TWO SPAN, CONTINUOUS BEAM – UNIFORM LOAD OVER ONE SPAN ONLY

R1 = V1 ........................................................ = 7 w16

R2 = V2 + V3................................................. = 5 w8

R3 = V3 ........................................................ = 1 w16

−

V2 ................................................................ = 9 w16

Mmax 7at x

16⎛ ⎞=⎜ ⎟⎝ ⎠

...................................... = 249 w512

M1 (at R2).................................................... = 2w

16

Mx (when x < ).......................................... = wx (7 8x)16

−



DESIGN INFORMATION Design Aid 11.1.4 Camber (Deflection) and Rotation Coefficients for Prestress Force and Loads

CAMBER END ROTATION

PRESTRESS PATTERN

EQUIVA-LENT

MOMENT OR LOAD

EQUIVALENT LOADING

(1)

M = Pe 2M

16El+ M

3El+ M

6El−

(2)

M = Pe 2M

16El+ M

6El+ M

3El−

(3)

M = Pe 2M

8El+ M

2El M

2El−

(4)

4PeN′

=3N

48El+

2N16El

2N

16El−

(5)

PeNb

′=

2 3b(3 4b )N24El

−

2b(1 b)N2El

− 2b(1 b)N2El−−

(6)

28Pew

′=

45w384El

3w

24El

3w24El

−



DESIGN INFORMATION Design Aid 11.1.4 Camber (Deflection) and Rotation Coefficients for Prestress Force and Loads (Cont.)

CAMBER END ROTATION

PRESTRESS PATTERN

EQUIVA-LENT

MOMENT OR LOAD

EQUIVALENT LOADING

(7)

w = 28Pe′l

45w768El

l 39w

384Ell

37w384El

− l

(8)

w = 28Pe′l

45w

768Ell

37w384El

l 39w

384El− l

(9)

w =

24Pe

(0.5 b)′

− l

w1 = w (0.5 b)b

−

25 b (3 2b )8 2⎡ ⎤− −⎢ ⎥⎣ ⎦

4w48Ell

3(1 b)(1 2b)w24El

− − l

3(1 b)(1 2b)w24El

− −− l

(10)

w =

24Pe

(0.5 b)′

− l

w1 = w (0.5 b)b

−

25 b (3 2b )16 4⎡ ⎤− −⎢ ⎥⎣ ⎦

4w

48Ell

29 b(2 b)8⎡ ⎤− −⎢ ⎥⎣ ⎦

3w48Ell

27 b(2 b)8

⎡ ⎤− + −⎢ ⎥⎣ ⎦

3w48Ell

(11)

w =

24Pe

(0.5 b)′

− l

w1 = w (0.5 b)b

−

25 b (3 2b )16 4⎡ ⎤− −⎢ ⎥⎣ ⎦

4w

48Ell

27 b(2 b)8⎡ ⎤− −⎢ ⎥⎣ ⎦

3w48Ell

29 b(2 b)8

⎡ ⎤− + −⎢ ⎥⎣ ⎦

3w48Ell

Determination of camber along length of member based on camber at midspan:

Camber at midspan = yc yx =

2

c c 2

x2y y

2

⎛ ⎞−⎜ ⎟⎝ ⎠−⎛ ⎞⎜ ⎟⎝ ⎠

l

l



DESIGN INFORMATION Design Aid 11.1.5 Moments in Beams with Fixed Ends

LOADING MOMENT AT A MOMENT AT CENTER MOMENT AT B

(1)

P8

− P8

P8

−

(2)

2P a(1 a)− − 2P a (1 a)− −

(3)

2P9

− P9

2P9

−

(4)

5P16

− 3P16

5P16

−

(5)

W12

− W24

W12

−

(6)

2W (1 2a 2a )12

− + − 2W (1 2a 2a )

24+ −

2W (1 2a 2a )12

− + −

(7)

2W (3a 2a )12

− − 2W a

6

2W (3a 2a )12

− −

(8)

2W a(6 8a 3a )12

− − + 2W a (4 3a)12

− −

(9)

5W48

− 3W48

5W48

−

(10)

W10

− W15

−

W = Total load on the beam.



DESIGN INFORMATION Design Aid 11.1.6 Torsion Diagrams, Reactions, and Rotations

(1)

At support: T = T θ = T

TGJ

(2)

At support: T = t θ =2

T

T2GJ

(3)

a: 1a

T bT = θ1 = 1

T

T abGJ

When a = b = 2

b: 1b

T aT = θ1 =T

T4GJ

(4)

a: 1 2a

T (b c) T cT + += θ1 = a

T

T aGJ

b: 2 1b

T c T aT −= θ2 = c

T

T cGJ

When a = b = c = T/8 T1 = T2 = T/2

c: 1 2c

T a T (a b)T + += and 1 2θ = θ = T

T6GJ

(5)

a: 1 2 3a

T (b c d) T (c d) T dT + + + + += θ2 = b a

T

T b T aGJ

+

b: 1 2 3b

T a T (c d) T dT − + + += θ1 = a

T

T aGJ

c: 1 2 3c

T a T (a b) T dT − − + += θ3 = d

T

T dGJ

d: 1 2 3d

T a T (a b) T (a b c)T − − − + + +=

(6)

Tsupport = t2

CLθ = 2

T

t8GJ

}G = Shear modulus See Section 6.6J = Torsion constant



DESIGN INFORMATION Design Aid 11.1.7 Moving Load Placement for Maximum Moment and Sheara

(1) SIMPLE BEAM – ONE CONCENTRATED MOVING LOAD R1 max = V1 max (AT x = 0) ..........................................= P

M max (at point of load, when x = 2

) ........................= P4

(2) SIMPLE BEAM – TWO EQUAL CONCENTRATED MOVING LOADS

R1 max = V1 max (AT x = 0) ...........................................= aP 2⎛ ⎞−⎜ ⎟⎝ ⎠

when a < (2 2)− ...................................................= 0.586

Mmax = under load 1, at x = 1 a2 2⎛ ⎞−⎜ ⎟⎝ ⎠

.......................= 2P a

2 2⎛ ⎞−⎜ ⎟⎝ ⎠

when a > (2 2)− ...............................................= 0.586

Mmax = with one load at center of span .....................= P4

(3) SIMPLE BEAM – TWO UNEQUAL CONCENTRATED MOVING LOADS

R1 max = V1 max (at x = 0) .............................................= 1 2aP P −+

Mmax under P1, at 2

1 2

P a1x2 P P⎛ ⎞

= −⎜ ⎟+⎝ ⎠.......................=

2

1 2x(P P )+

Mmax may occur with larger load at center

of span and other load off span ................................= 1P4

a. Souce “Manual of Steel Construction, Allowable Stress Design,” Ninth Edition, 1989 American Institute of Steel Construction, Chicago, IL.



DESIGN INFORMATION Design Aid 11.1.8 Moments, Shears, and Deflections in Beams with Overhangs

LOADING AND SUPPORT

REACTIONS AND

VERTICAL SHEAR

BENDING MOMENT M AND MAXIMUM

BENDING MOMENT

DEFLECTION y, MAXIMUM DEFLECTION, AND END SLOPE θ

EQUAL OVERHANGS, UNIFORM LOAD

RB = RC = W2

(A to B) V =

W(c x)−−l

(B to C) V =

1 x cW2

+⎛ ⎞−⎜ ⎟⎝ ⎠l

(C to D) V = W(c d x)+ −

l

(A to B)M = 2W (c x)2

− −l

(B to C)M = 2W [(c x(d x)]

2− − −

l

M = 2Wc

2−

l at B and C

M = 2

2W dc2 4

⎛ ⎞⎜ ⎟− −⎜ ⎟⎝ ⎠l

dat x2

= if d > 2c, M = 0

at x = 2

2d d c2 4

± −

if c = 0.207 l ,

M = W46.62

− l at x = 0 = d

and M = W46.62

l

ar x = d2

x is considered positive on both sides of the origin.

(A to B) y = 2 2 3Wx [6c (d x) x (4c x) d ]24El

− + − − −l

(B to C) y = 2 2Wx(d x) [x(d x) d 6c ]24El

−− − + −l

y = 2 3Wc [3c (c 2d) d ]24El

− + −l

at A and D

y = 2

2 2Wd d(5d 24c ) AT x384El 2

− − =l

if 2c < d < 2.449c, the maximum deflection between supports is:

y = 2

2 2 2 2W d d(6c d ) AT x 3 c96El 2 4

⎛ ⎞⎜ ⎟− = ± −⎜ ⎟⎝ ⎠l

θ = 2 3 3W (6c d 4c d ) AT A24El

+ −l

θ = 2 3 3W (6c d 4c d ) AT D24El

− + −l

UNEQUAL OVERHANGS, UNIFORM LOAD

RB = W (c d e)2d

+ −

RC = W (d e c)2d

+ −

(A to B) V = W (c x)− −l

(B to C) V =

BWR (c x)− +l

(C to D) V =

W (d e x)− + −l

(A to B) M = 2W (c x)

2− −

l

(B to C) M = 2

BW (c x) R x2

− − +l

(C to D) M = 2W (e d x)

2− + −

l

M = 2Wc

2−

l at B

M = 2We

2−

l at C

Mmax between supports

= 2 21

W (c x )2

−l

at x = x1

= 2 2 2c d e

2d+ − if x1 > c,

M = 0

AT x = 2 21 1x x c± −

x is considered positive on both sides of the origin.

(A to B) y = Wx24El

−l

2 2 2 2 3[2d(e 2c ) 6c x x (4c x) d ]+ + − − −

(B to C) y = Wx(d x)24El

−−l

2 2 2 2 22x(d x) d 2(c e ) [e x c (d x)]d

⎧ ⎫− + − + − + −⎨ ⎬⎩ ⎭

(C to D) y = 2 2 2W(x d) [2d(c 2e ) 6e (x d)24El

−− + + −l

2 3(x d) (4e d x) d ]− − + − −

y = 2 2 3 3Wc [2d(e 2c ) 3c d ]24El

− + + −l

AT A

y = 2 2 3 3We [2d(c 2e ) 3e d ]24El

− + + −l

AT D

This case is too complicated to obtain a general expression for critical deflections between the supports.

θ = 3 2 3 2W (4c 4c d d 2de )24El

+ − +l

AT A

θ = 2 2 3 3W (2c d 4de d 4e )24El

− + − +l

AT D



11.2 MATERIAL PROPERTIES CONCRETE

Design Aid 11.2.1 Table of Concrete Stresses (psi)

cf ′ c0.45f′ c0.6f ′ cf ′ c2 f ′ c3.5 f ′ c4 f ′ c5 f ′ c6 f ′ c7.5 f ′ c12 f ′

2500 1125 1500 50 100 175 200 250 300 375 600 3000 1350 1800 55 110 192 219 274 329 411 657 3500 1575 2100 59 118 207 237 296 355 444 710 4000 1800 2400 63 126 221 253 316 379 474 759 5000 2250 3000 71 141 247 283 354 424 530 849 6000 2700 3600 77 155 271 310 387 465 581 930 7000 3150 4200 84 167 293 335 418 502 627 1004 8000 3600 4800 89 179 313 358 447 537 671 1073 9000 4050 5400 95 190 332 379 474 569 712 1138

10000 4500 6000 100 200 350 400 500 600 750 1200

Design Aid 11.2.2 Concrete Modulus of Elasticity as Affected by Unit Weight (pcf) and Strength (psi)



MATERIAL PROPERTIES PRESTRESSING STEEL

Design Aid 11.2.3 Properties and Design Strengths of Prestressing Strand and Wire

Seven-Wire Strand, fpu = 270 ksi

Nominal Diameter, in. E/K U/AH Q/S Q/S Speciala O/AH E/G

Area, sq in. 0.085 0.115 0.153 0.167 0.192 0.217

Weight, plf 0.29 0.40 0.52 0.53 0.65 0.74

0.7fpuAps, kips 16.1 21.7 28.9 31.6 36.3 41.0

0.75fpuAps, kips 17.2 23.3 31.0 33.8 38.9 43.0

0.8fpuAps, kips 18.4 24.8 33.0 36.1 41.5 46.9

fpuAps, kips 23.0 31.1 41.3 45.1 51.8 58.6

Seven-Wire Strand, fpu = 250 ksi

Nominal Diameter, in. Q/F T/AH E/K U/AH Q/S E/G

Area, sq in. 0.036 0.058 0.080 0.108 0.144 0.216

Weight, plf 0.12 0.20 0.27 0.37 0.49 0.74

0.7fpuAps, kips 6.3 10.2 14.0 18.9 25.2 37.8

0.8fpuAps, kips 7.2 11.6 16.0 21.6 28.8 43.2

fpuAps, kips 9.0 14.5 20.0 27.0 36.0 54.0

Three- and Four-Wire Strand, fpu = 250 ksi

Nominal Diameter, in. Q/F T/AH E/K U/AH

No. of wires 3 3 3 4

Area, sq in. 0.036 0.058 0.075 0.106

Weight, plf 0.12 0.20 0.26 0.36

0.7fpuAps, kips 6.3 10.2 13.1 18.6

0.8fpuAps, kips 7.2 11.6 15.0 21.2

fpuAps, kips 9.0 14.5 18.8 26.5

Prestressing Wire

Diameter 0.105 0.120 0.135 0.148 0.162 0.177 0.192 0.196 0.250 0.276

Area, sq in. 0.0087 0.0114 0.0143 0.0173 0.0206 0.0246 0.0289 0.0302 0.0491 0.0598

Weight, plf 0.030 0.039 0.049 0.059 0.070 0.083 0.098 0.10 0.17 0.20

Ult. strength, fpu ksi 279 273 268 263 259 255 250 250 240 235

0.7fpuAps, kips 1.70 2.18 2.68 3.18 3.73 4.39 5.06 5.29 8.25 9.84

0.8fpuAps, kips 1.94 2.49 3.07 3.64 4.27 5.02 5.78 6.04 9.43 11.24

fpuAps, kips 2.43 3.11 3.83 4.55 5.34 6.27 7.23 7.55 11.78 14.05 a. The Q/S in. special strand has a larger actual diameter than the Q/S in. regular strand. The table values take this difference into account.




Design Aid 11.2.4 Properties and Design Strengths of Prestressing Bars

Plain Prestressing Bars, fpu = 145 ksia

Nominal Diameter, in. E/K U/K 1 1Q/K 1Q/F 1E/K

Area, sq in. 0.442 0.601 0.785 0.994 1.227 1.485

Weight, plf 1.50 2.04 2.67 3.38 4.17 5.05

0.7fpuAps, kips 44.9 61.0 79.7 100.9 124.5 150.7

0.8fpuAps, kips 51.3 69.7 91.0 115.3 142.3 172.2

fpuAps, kips 64.1 87.1 113.8 144.1 177.9 215.3

Plain Prestressing Bars, fpu = 160 ksia

Nominal Diameter, in. E/K U/K 1 1Q/K 1Q/F 1E/K

Area, sq in. 0.442 0.601 0.785 0.994 1.227 1.485

Weight, plf 1.50 2.04 2.67 3.38 4.17 5.05

0.7fpuAps, kips 49.5 67.3 87.9 111.3 137.4 166.3

0.8fpuAps, kips 56.6 77.0 100.5 127.2 157.0 190.1

fpuAps, kips 70.7 96.2 125.6 159.0 196.3 237.6

Deformed Prestressing Bars

Nominal Diameter, in. T/K 1 1 1Q/F 1Q/F 1E/K

Area, sq in. 0.28 0.85 0.85 1.25 1.25 1.58

Weight, plf 0.98 3.01 3.01 4.39 4.39 5.56

Ult. strength, fpu, ksi 157 150 160a 150 160a 150

0.7fpuAps, kips 30.5 89.3 95.2 131.3 140.0 165.9

0.8fpuAps, kips 34.8 102.0 108.8 150.0 160.0 189.6

fpuAps, kips 43.5 127.5 136.0 187.5 200.0 237.0

Stress-strain characteristics (all prestressing bars):

For design purposes, the following assumptions are satisfactory:

Es = 29,000 ksi

fy = 0.95fpu

a. Verify availability before specifying.




Design Aid 11.2.5 Typical Stress-Strain Curve, 7-Wire Low-Relaxation Prestressing Strand

Note: approximate strain at rupture is 0.05 to 0.07 in./in. These curves can be approximated by the following equations: 250 ksi strand: 270 ksi strand: ps 0.0076ε ≤ : ps psf 28,500 (ksi)= ε ps 0.0086ε ≤ : ps psf 28,500 (ksi)= ε

ps 0.0076ε > : psps

0.04f 250 (ksi)0.0064

= −ε −

ps 0.0086ε > : psps

0.04f 270 (ksi)0.007

= −ε −




Design Aid 11.2.6 Transfer and Development Lengths for 7-Wire Uncoated Strand

The ACI 318-02 (Section 12.9.1) equation for required development length may be rewritten as: d = se b ps se b(f / 3)d (f f )d+ −

where: d = required development length, in. fse = effective prestress, ksi fps = stress in prestressing steel at nominal strength, ksi db = nominal diameter of strand, in. The first term in the equation is the transfer length and the second term is the additional length required for the stress increase (fps – fse) corresponding to the nominal strength.

Transfer and development lengtha in inches fse = 150 ksi fse = 160 ksi fse = 170 ksi

Development Length Development Length Development Length

fps, ksi fps, ksi fps, ksi

Nominal Diameter,

in. Trans-

fer Length 240 250 260 270

Trans-fer

Length 240 250 260 270

Trans-fer

Length 240 250 260 270E/K 18.8 52.5 56.3 60.0 63.8 20.0 50.0 53.8 57.5 61.3 21.3 47.5 51.3 55.0 58.8U/AH 21.9 61.3 65.6 70.0 74.4 23.3 58.3 62.7 67.0 71.4 24.8 55.4 59.8 64.2 68.5Q/S 25.0 70.0 75.0 80.0 85.0 26.7 66.7 71.7 76.7 81.7 28.3 66.3 68.3 73.3 78.3

Q/S Sb 26.1 73.1 78.4 83.6 88.8 27.9 69.7 74.9 80.1 85.4 29.6 66.1 71.3 76.6 81.8O/AH 28.1 78.8 84.4 90.0 95.6 30.0 75.0 80.6 86.3 91.9 31.9 71.3 76.9 82.5 88.1E/G 30.0 84.0 90.0 96.0 102.0 32.0 80.0 86.0 92.0 98.0 34.0 76.0 82.0 88.0 94.0

a. The development length values given in the table must be doubled where bonding of the strand does not extend to the member end and

the member is designed such that tension in the precompressed tensile zone is produced under service loads (see ACI 318-02, Section 12.9.3).

b. The Q/S in. special (Q/S S) strand has a larger nominal diameter than the Q/S in. regular (Q/S) strand. The table values for transfer and development length reflect this difference in diameters.



MATERIAL PROPERTIES REINFORCING BARS

Design Aid 11.2.7 Reinforcing Bar Data

ASTM STANDARD REINFORCING BARS NOMINAL DIMENSIONS BAR SIZEa

DESIGNATION DIAMETER AREA WEIGHT OR MASS U.S. CUSTOMARY SI in. mm in.2 mm2 lb/ft kg/m

#3 #10 0.375 9.5 0.11 71 0.376 0.560 #4 #13 0.500 12.7 0.20 129 0.668 0.994 #5 #16 0.625 15.9 0.31 199 1.043 1.552 #6 #19 0.750 19.1 0.44 284 1.502 2.235 #7 #22 0.875 22.2 0.60 387 2.044 3.042 #8 #25 1.000 25.4 0.79 510 2.670 3.973 #9 #29 1.128 28.7 1.00 645 3.400 5.060 #10 #32 1.270 32.3 1.27 819 4.303 6.404 #11 #36 1.410 35.8 1.56 1006 5.313 7.907 #14 #43 1.693 43.0 2.25 1452 7.650 11.380 #18 #57 2.257 57.3 4.00 2581 13.600 20.240

a. Many mills will mark and supply bars only with metric (SI) designation, which is a soft conversion. Soft conversion means that the metric

(SI) bars have exactly the same dimensions and properties as the equivalent U.S. customary designation.

STANDARD HOOKS STIRRUP AND TIE-HOOKS

180° 90° 90° 135° BAR SIZE D

A OR G J A OR G D

A OR G A OR G H U.S. SI U.S. SI U.S. SI U.S. SI U.S. SI U.S. SI U.S. SI U.S. SI U.S. SI #3 #10 2Q/F 60 5 125 3 80 6 150 1Q/S 40 4 105 4 105 2Q/S 65 #4 #13 3 80 6 150 4 105 8 200 2 50 4Q/S 115 4Q/S 115 3 80 #5 #16 3E/F 95 7 175 5 130 10 250 2Q/S 65 6 155 5Q/S 140 3E/F 95 #6 #19 4Q/S 115 8 200 6 155 1–0 300 4Q/S 115 1–0 305 8 205 4Q/S 115 #7 #22 5Q/F

a 135a 10 250 7 180 1–2 375 5Q/Fa 135a 1–2 355 9 230 5Q/F 135

#8 #25 6a 155a 11 275 8 205 1–4 425 6a 155a 1–4 410 10Q/S 270 6 155 #9 #29 9Q/S 240 1–3 375 11E/F 300 1–7 475

#10 #32 10E/F 275 1–5 425 1–1Q/F 335 1–10 550 #11 #36 12 305 1–7 475 1–2E/F 375 2–0 600 #14 #43 18Q/F 465 2–3 675 1–9E/F 550 2–7 775 #18 #57 24 610 3–0 925 2–4Q/S 725 3–5 1050

U.S. CUSTOMARY UNITS: in. or ft-in. SI UNITS: mm

a. ASTM A767 requires that bars bent cold prior to hot dip galvanizing must be fabricated to a minimum bend diameter equal to 7 in. for #7

bar and 8 in. for #8 bar.




Design Aid 11.2.8 Location of Reinforcement Confined by Stirrups or Ties

Z Dimension (in.)

Stirrup or Tie Size

#3 #4 #5

#4 U/K 1Q/AH 1Q/F

#5 U/K 1Q/K 1T/AH

#6 QT/AH 1E/AH 1E/K

#7 1 1E/AH 1U/AH

#8 1Q/AH 1Q/F 1U/AH

#9 1Q/AH 1T/AH 1Q/S

#10 1Q/K 1T/AH 1Q/S

Mai

n R

einf

orce

men

t Siz

e

#11 1E/AH 1E/K 1O/AH To determine location of main reinforcement, add specified cover to the “z” dimension from above table.




Design Aid 11.2.9 Required Development Lengthsa for Reinforcing Bars (Grade 60)

Tension Development Length:

d = b

c

d2400

f ′; min. 12 in. (#6 and smaller)

d = b

c

d3000

f ′; min. 12 in. (#7 and larger)

(Note: for Grade 40 bars, replace 2400 and 3000 with 1600 and 2000, respectively.) Multiply d values by : (a) 1.3 for lightweight concrete (b) 1.3 for “top bars” (c) 1.5 for epoxy coated bars with cover < 3db or

clear spacing < 6db, otherwise multiply by 1.2 (Note: Product of factors (b) and (c) need not exceed

1.7) (d) 1.5 for bars with less than minimum stirrups or

ties, clear spacing less than 2db or clear cover less than db.

(e) As (required)/As (provided) for excess reinforce-ment unless development of fy is specially required. This multiplier is not to be applied to lap splices per ACI 318-02, Section R12.15.1.

Compression Development Length:

d = b

c

1200df ′

; min. 18db and 8 in.

(Note: For Grade 40 bars, replace 1200 with 800 and 18 with 12) Multiply d values by: (a) As(required)/As (provided) for excess reinforcement (b) 0.75 for adequate spiral or tie enclosure (see ACI 318-02, Section 12.3.3b) Compression Splice Lap Length: Lap length = 30db; min. 12 in. The values of cf ′ used in these equations shall not exceed 100 psi (see Section 12.1.2, ACI 318-02)

cf ′ = 3000 psi cf′ = 4000 psi cf′ = 5000 psi cf ′ = 6000 psi

Tension Com-pres-sion




Bar Size

d 1.3 d 1.5 d d d 1.3 d 1.5 d d d 1.3 d 1.5 d d d 1.3 d 1.5 d d

Min. Comp. Splice

3 16 21 25 8 14 18 21 8 13 17 19 8 12 15 17 8 12 4 22 28 33 11 19 25 28 9 17 22 25 9 15 20 23 9 15 5 27 36 41 14 24 31 36 12 21 28 32 11 19 25 29 11 19 6 33 43 49 16 28 37 43 14 25 33 38 14 23 30 35 14 23 7 48 62 72 19 42 54 62 17 37 48 56 16 34 44 51 16 26 8 55 71 82 22 47 62 71 19 42 55 64 18 39 50 58 18 30 9 62 80 93 25 54 70 80 21 48 62 72 20 44 57 66 20 34 10 70 90 104 28 60 78 90 24 54 70 81 23 49 64 74 23 38 11 77 100 116 31 67 87 100 27 60 78 90 25 55 71 82 25 42

cf ′ = 7000 psi cf′ = 8000 psi cf′ = 9000 psi cf ′ = 10,000 psi





Bar Size

d 1.3 d 1.5 d d d 1.3 d 1.5 d d d 1.3 d 1.5 d d d 1.3 d 1.5 d d

Min. Comp. Splice

3 12 14 16 8 12 13 15 8 12 12 14 8 9 12 14 8 12 4 14 19 22 9 13 17 20 9 13 16 19 8 12 16 18 8 15 5 18 23 27 11 17 22 25 11 16 21 24 8 15 20 23 8 19 6 22 28 32 14 20 26 30 14 19 25 28 9 18 23 27 9 23 7 31 41 47 16 29 38 44 16 28 36 42 11 26 34 39 11 26 8 36 47 54 18 34 44 50 18 32 41 47 13 30 39 45 12 30 9 40 53 61 20 38 49 57 20 36 46 54 14 34 44 51 14 34 10 46 59 68 23 43 55 64 23 40 52 60 16 38 49 56 15 38 11 51 66 76 25 47 61 71 25 45 58 67 17 41 54 62 17 42

a. For limitations and items related to hooked bars, stirrups or ties in excess of minimum, and spacing of non-contact lap splices, etc., see

ACI 318-02, Chapter 12.




Design Aid 11.2.9 Required Development Lengths for Reinforcing Bars (Grade 60) (Cont.)

Minimum tension embedment lengths, dh , for standard hooks, in. General use (non-seismic) [see ACI 318-02, Section 12.5.2 and 12.5.3(a)]

Normal weight concrete, cf ′ (psi) Bar Size 3,000 4,000 5,000 6,000 7,000 8,000 9,000 10,000

3 4 5

6 8

10

6 7 9

6 6 8

6 6 7

6 6 7

6 6 6

6 6 6

6 6 6

6 7 8

12 14 16

10 12 14

9 11 12

8 10 11

8 9

10

7 9

10

7 8 9

6 7 8

9 10 11

18 20 22

15 17 19

14 15 17

13 14 16

12 13 14

11 12 14

10 11 13

9 11 12

Notes: 1. Side Cover ≥ 2Q/S in. 2. End Cover (90° hooks) ≥ 2 in.

Minimum tension embedment lengths, dh , for standard hooks, in. Special confinement (non-seimic) [see ACI 318-02, Section 12.5.3(b)]

Normal weight concrete, cf ′ (psi) Bar Size 3,000 4,000 5,000 6,000 7,000 8,000 9,000 10,000

3 4 5

6 6 8

6 6 7

6 6 6

6 6 6

6 6 6

6 6 6

6 6 6

6 6 6

6 7 8

10 11 13

8 10 11

7 9

10

7 8 9

6 7 8

6 7 8

6 6 7

6 6 6

9 10 11

14 16 18

12 14 15

11 12 14

10 11 13

9 11 12

9 10 12

8 9

10

7 9

10 Notes: 1. Side Cover ≥ 2Q/S in. 2. End Cover (90° hooks) ≥ 2 in. BARS WITH STANDARD HOOKS

Standard 180° Hook

Standard 90° Hook

Dimension a = 4db for #3 through #8, = 5db for #9, #10 and #11 Modification factors: Grade 40 bars = 0.67 Lightweight concrete = 1.3 Epoxy coated reinforcement = 1.2



MATERIAL PROPERTIES STRUCTURAL WELDED WIRE REINFORCEMENT (WWR)

Design Aid 11.2.10 Common Styles of Structural Welded Wire Reinforcementa

Style designation Steel area in.2 per ft Former designation

(By steel wire gage) Current Designation

(By W-number) Longit. Trans.

Approximate Weight

lb per 100 ft2 6x6–10x10 4x12–8x10c 6x6–8x8 4x4–10x10

6x6–W1.4xW1.4 4x12–W2.1xW1.4 6x6–W2.1xW2.1 4x4–W1.4xW1.4

0.029 0.062 0.041 0.043

0.029 0.014 0.041 0.043

21 26 30 31

4x12–7x10c 6x6–6x6b 4x4–8x8 6x6–4x4b

4x12–W2.5xW1.4 6x6–W2.9xW2.9 4x4–W2.1xW2.1 6x6–W4.0xW4.0

0.074 0.058 0.062 0.080

0.014 0.058 0.062 0.080

31 42 44 58

4x4–6x6 6x6–2x2b 4x4–4x4b 4x4–3x3b 4x4–2x2b

4x4–W2.9xW2.9 6x6–W5.5xW5.5d 4x4–W4.0xW4.0 4x4–W4.7xW4.7 4x4–W5.5xW5.5d

0.087 0.110 0.120 0.141 0.165

0.087 0.110 0.120 0.141 0.165

62 80 85 102 119

a. Source: Manual of Standard Practice – Structural Welded Wire Reinforcement, Wire Reinforcement Institute, 1992, Findlay, Ohio. b. Commonly available in 8 ft x 12 ft or 8 ft x 15 ft sheets. c. These items may be carried in sheets by various manufacturers in certain parts of the U.S. and Canada. d. Exact W-number size for 2 gage is 5.4.



MATERIAL PROPERTIES STRUCTURAL WELDED WIRE REINFORCEMENT (WWR)

Design Aid 11.2.11 Wires Used in Structural Welded Wire Reinforcementa

Area – in.2 per ft of width Wire Size Number Center to Center Spacing, in.

Plainb Deformedc

Nominal Diameter

in.

Nominal Weight

plf 2 3 4 6 8 10 12 W45 D45 0.757 1.530 2.700 1.800 1.350 0.900 0.675 0.540 0.450 W31 D31 0.628 1.054 1.860 1.240 0.930 0.620 0.465 0.372 0.310 W30 D30 0.618 1.020 1.800 1.200 0.900 0.600 0.450 0.360 0.300 W28 D28 0.597 0.952 1.680 1.120 0.840 0.560 0.420 0.336 0.280 W26 D26 0.575 0.934 1.560 1.040 0.780 0.520 0.390 0.312 0.260 W24 D24 0.553 0.816 1.440 0.960 0.720 0.480 0.360 0.288 0.240 W22 D22 0.529 0.748 1.320 0.880 0.660 0.440 0.330 0.264 0.220 W20 D20 0.504 0.680 1.200 0.800 0.600 0.400 0.300 0.240 0.200 W18 D18 0.478 0.612 1.080 0.720 0.540 0.360 0.270 0.216 0.180 W16 D16 0.451 0.544 0.960 0.640 0.480 0.320 0.240 0.192 0.160 W14 D14 0.422 0.476 0.840 0.560 0.420 0.280 0.210 0.168 0.140 W12 D12 0.390 0.408 0.720 0.480 0.360 0.240 0.180 0.144 0.120 W11 D11 0.374 0.374 0.660 0.440 0.330 0.220 0.165 0.132 0.110 W10.5 D10.5 0.366 0.357 0.630 0.420 0.315 0.210 0.157 0.126 0.105 W10 D10 0.356 0.340 0.600 0.400 0.300 0.200 0.150 0.120 0.100 W9.5 D9.5 0.348 0.323 0.570 0.380 0.285 0.190 0.143 0.114 0.095 W9 D9 0.338 0.306 0.540 0.360 0.270 0.180 0.135 0.108 0.090 W8.5 D8.5 0.329 0.289 0.510 0.340 0.255 0.170 0.128 0.102 0.085 W8 D8 0.319 0.272 0.480 0.320 0.240 0.160 0.120 0.096 0.080 W7.5 D7.5 0.309 0.255 0.450 0.300 0.225 0.150 0.113 0.090 0.075 W7 D7 0.298 0.238 0.420 0.280 0.210 0.140 0.105 0.084 0.070 W6.5 D6.5 0.288 0.221 0.390 0.260 0.195 0.130 0.098 0.078 0.065 W6 D6 0.276 0.204 0.360 0.240 0.180 0.120 0.090 0.072 0.060 W5.5 D5.5 0.264 0.187 0.330 0.220 0.165 0.110 0.083 0.066 0.055 W5 D5 0.252 0.170 0.300 0.200 0.150 0.100 0.075 0.060 0.050 W4.5 D4.5 0.240 0.153 0.270 0.180 0.135 0.090 0.068 0.054 0.045 W4 D4 0.225 0.136 0.240 0.160 0.120 0.080 0.060 0.048 0.040 W3.5 0.211 0.119 0.210 0.140 0.105 0.070 0.053 0.042 0.035 W3 0.195 0.102 0.180 0.120 0.090 0.060 0.045 0.036 0.030 W2.9 0.192 0.098 0.174 0.116 0.087 0.058 0.044 0.035 0.029 W2.5 0.178 0.085 0.150 0.100 0.075 0.050 0.038 0.030 0.025 W2.1 0.162 0.070 0.126 0.084 0.063 0.042 0.032 0.025 0.021 W2 0.159 0.068 0.120 0.080 0.060 0.040 0.030 0.024 0.020 W1.5 0.138 0.051 0.090 0.060 0.045 0.030 0.022 0.018 0.015 W1.4 0.135 0.049 0.084 0.056 0.042 0.028 0.021 0.017 0.014

a. Source: Manual of Standard Practice—Structural Welded Wire Reinforcement, Wire Reinforcement Institute, 1992, Findlay, Ohio. b. ASTM A 82, Available fy = 65,000 psi to 80,000 psi in 2500 psi increments. c. ASTM A 496, Available fy = 70,000 psi to 80,000 psi in 2500 psi increments.



MATERIAL PROPERTIES BAR AREA EQUIVALENTS

Design Aid 11.2.12 Bar Area Equivalents in a One Foot Wide Section

Bar Reinforcing Bar Size (Nominal Diameter – in.) Area

Spacing c/c (in.)

#3 (0.375)

#4 (0.500)

#5 (0.625)

#6 (0.750)

#7 (0.875)

#8 (1.000)

#9 (1.128)

#10 (1.270)

#11 (1.410) Range

2 0.66 1.20 1.86 2.64 3.60 4.74 Exceeds min. bar clear 2Q/S 0.53 0.96 1.49 2.11 2.88 3.79 4.80 spacing of db 3 0.44 0.80 1.24 1.76 2.40 3.16 4.00 5.08 6.24

3Q/S 0.38 0.69 1.06 1.51 2.06 2.71 3.43 4.35 5.35 4 0.33 0.60 0.93 1.32 1.80 2.37 3.00 3.81 4.68

4Q/S 0.29 0.53 0.83 1.17 1.60 2.11 2.67 3.39 4.16 5 0.26 0.48 0.74 1.06 1.44 1.90 2.40 3.05 3.74

5Q/S 0.24 0.44 0.68 0.96 1.31 1.72 2.18 2.77 3.40 6 0.22 0.40 0.62 0.88 1.20 1.58 2.00 2.54 3.12

≥ 3.0 sq in.

6Q/S 0.20 0.37 0.57 0.81 1.11 1.46 1.85 2.34 2.88 7 0.19 0.34 0.53 0.75 1.03 1.35 1.71 2.18 2.67

7Q/S 0.18 0.32 0.50 0.70 0.96 1.26 1.60 2.03 2.50 8 0.17 0.30 0.47 0.66 0.90 1.19 1.50 1.91 2.34

8Q/S 0.16 0.28 0.44 0.62 0.85 1.12 1.41 1.79 2.20 9 0.15 0.27 0.41 0.59 0.80 1.05 1.33 1.69 2.08

2.0 to 3.0 sq in.

9Q/S 0.14 0.25 0.39 0.56 0.76 1.00 1.26 1.60 1.97 10 0.13 0.24 0.37 0.53 0.72 0.95 1.20 1.52 1.87

10Q/S 0.13 0.23 0.35 0.50 0.69 0.90 1.14 1.45 1.78 11 0.12 0.22 0.34 0.48 0.65 0.86 1.09 1.39 1.70

11Q/S 0.11 0.21 0.32 0.46 0.63 0.82 1.04 1.33 1.63 12 0.11 0.20 0.31 0.44 0.60 0.79 1.00 1.27 1.56

1.5 to 2.0 sq in.

13 0.10 0.18 0.29 0.41 0.55 0.73 0.92 1.17 1.44 14 0.09 0.17 0.27 0.38 0.51 0.68 0.86 1.09 1.34 15 0.09 0.16 0.25 0.35 0.48 0.63 0.80 1.02 1.25 16 0.08 0.15 0.23 0.33 0.45 0.59 0.75 0.95 1.17 17 0.08 0.14 0.22 0.31 0.42 0.56 0.71 0.90 1.10 18 0.07 0.13 0.21 0.29 0.40 0.53 0.67 0.85 1.04

1.0 to 1.5 sq in.

19 0.07 0.13 0.20 0.28 0.38 0.50 0.63 0.80 0.99 20 0.07 0.12 0.19 0.26 0.36 0.47 0.60 0.76 0.94 21 0.06 0.11 0.18 0.25 0.34 0.45 0.57 0.73 0.89 22 0.06 0.11 0.17 0.24 0.33 0.43 0.55 0.69 0.85 23 0.06 0.10 0.16 0.23 0.31 0.41 0.52 0.66 0.81 24 0.06 0.10 0.16 0.22 0.30 0.40 0.50 0.64 0.78

0.75 to 1.0 sq in.

Area Range ≤ 0.25 sq in. 0.25 to 0.50 sq in. 0.50 to 0.75 sq in.

NOTE: Check minimum requirements for temperature and shrinkage steel.

How to use this design aid. Given a design (or minimum temperature/shrinkage) reinforcement are required per foot, enter the design aid along right column or bottom row. Select one of the bar area ranges from that given sq in./ft, and follow the range band upward and/or to the left. Select the combina-tion of bar size and spacing satisfying the design and spacing requirements for the section.

Example. A design that requires reinforcement at 0.62 in.2/ft, with bar size restricted to No. 7 or smaller. Enter the design aid in the 0.50 to 0.75 sq in. range located along the bottom row. Follow the shaded band up to the top of the table. Select one of the following combinations: one layer of No.4 at about 3.5 in. o.c., No. 5 at 6 in. o.c., No. 6 at 8.5 in. o.c., or No. 7 at 11.5 in. o.c. Similar spacing(s) could be determined for two reinforcement layers, if desired.



MATERIALS PROPERTIES Design Aid 11.2.13 ACI Required Minimum Reinforcement Areas Per Foot Width of Section

Member Thickness or Vertical Height – h (in.) Element/Category As/Agross

Ratioa 2 3 4 5 6 7 8 9 Structural Slabs

fy < 60 ksi 0.0020 0.048 0.072 0.096 0.120 0.144 0.168 0.192 0.216

fy = 60 ksi 0.0018 0.043 0.065 0.086 0.108 0.130 0.151 0.173 0.194

fy = 70 ksi 0.0015 0.036 0.054 0.072 0.090 0.108 0.126 0.144 0.162

Minimumb 0.0014 0.034 0.050 0.067 0.084 0.101 0.118 0.134 0.151

Walls One layer permissible in walls

Verticalc 0.0012 0.029 0.043 0.058 0.072 0.086 0.101 0.115 0.130

Verticald 0.0015 0.036 0.054 0.072 0.090 0.108 0.126 0.144 0.162 Cast-in-place Horizontalc 0.0020 0.048 0.072 0.096 0.120 0.144 0.168 0.192 0.216

Horizontald 0.0025 0.060 0.090 0.120 0.150 0.180 0.210 0.240 0.270

Precast(5) Vert & Horiz 0.0010 0.024 0.036 0.048 0.060 0.072 0.084 0.096 0.108

Member thickness or Vertical Height – h (in.) Element/Category As/Agross

Ratioa 10 11 12 15 18 21 24 30 36 42 48

Structural Slabs

fy < 60 ksi 0.0020 0.240 0.264 0.288 0.360 0.432 0.504 0.546 0.720 0.864 1.008 1.152

fy = 60 ksi 0.0018 0.216 0.238 0.259 0.324 0.389 0.454 0.518 0.648 0.778 0.907 1.037

fy = 70 ksi 0.0015 0.180 0.198 0.216 0.270 0.324 0.378 0.432 0.540 0.648 0.756 0.864

Minimumb 0.0014 0.168 0.185 0.202 0.252 0.302 0.353 0.403 0.504 0.605 0.706 0.806

Walls Two layers required in walls

Verticalc 0.0012 0.144 0.158 0.173 0.216 0.259 0.302 0.346 0.432 0.518 0.605 0.691

Verticald 0.0015 0.180 0.198 0.216 0.270 0.324 0.378 0.432 0.540 0.648 0.756 0.864Cast-in-

place Horizontalc 0.0020 0.240 0.264 0.288 0.360 0.432 0.504 0.576 0.720 0.864 1.008 1.152

Horizontald 0.0025 0.300 0.330 0.360 0.450 0.540 0.630 0.720 0.900 1.080 1.260 1.440

Pre-cast(e) Vert. & Horiz. 0.0010 0.120 0.132 0.144 0.180 0.216 0.252 0.288 0.360 0.432 0.504 0.576

Notes: a. Ratio is reinforcement area to gross concrete area. b. Minimum controls for fy > 77 ksi. c. For deformed bars not larger than #5, with a specified yield strength not less than 60,000 psi,or for welded wire reinforcement (plain or

deformed) not larger than W31 or D31. d. For other deformed bars. e. Refer to ACI 318-02, Section 16.4.2.



11.3 STANDARD BOLTS, NUTS AND WASHERS Design Aid 11.3.1 Dimensions of Nuts and Bolts

Bolt Heads

Bolt head dimensions, rounded to nearest Q/AH inch, are in accordance with ANSI B18.2.1 – 1972 (Square and Hex) and ANSI 18.5 – 1971 (Countersunk)

Standard Dimensions for Bolt Heads

Square Hex Heavy Hex Countersunk Dia. of Bolt D Width

F Width

C Height

H Width

F Width

C Height

H Width

F Width

C Height

H Diam.

D Height

H in. in. in. in. in. in. in. in. in. in. in. in.

Q/F E/K Q/S E/AH U/AH Q/S E/AH ... ... ... Q/S Q/K E/K O/AH QE/AH Q/F O/AH T/K Q/F ... ... ... QQ/AH E/AH Q/S E/F 1Q/AH T/AH E/F U/K E/K U/K 1 E/K U/K Q/F T/K QT/AH 1T/AH U/AH QT/AH 1Q/AH U/AH 1Q/AH 1Q/F U/AH 1Q/K T/AH E/F 1Q/K 1O/AH Q/S 1Q/K 1T/AH Q/S 1Q/F 1U/AH Q/S 1E/K E/K U/K 1T/AH 1U/K T/K 1T/AH 1Q/S O/AH 1U/AH 1QQ/AH O/AH 1O/AH U/AH

1 1Q/S 2Q/K QQ/AH 1Q/S 1E/F QQ/AH 1T/K 1U/K QQ/AH 1QE/AH Q/S 1Q/K 1QQ/AH 2E/K E/F 1QQ/AH 1QT/AH E/F 1QE/AH 2Q/AH E/F 2Q/AH O/AH 1Q/F 1U/K 2T/K U/K 1U/K 2E/AH U/K 2 2T/AH U/K 2Q/F T/K 1E/K 2Q/AH 2QT/AH QT/AH 2Q/AH 2E/K QT/AH 2E/AH 2Q/S QT/AH 2Q/S QQ/AH 1Q/S 2Q/F 3E/AH 1 2Q/F 2T/K 1 2E/K 2E/F 1 2QQ/AH E/F 1E/F ... ... ... 2T/K 3 1E/AH 2E/F 3E/AH 1E/AH ... ...

2 ... ... ... 3 3U/AH 1E/K 3Q/K 3T/K 1E/K ... ...

2Q/F ... ... ... 3E/K 3U/K 1Q/S 3Q/S 4Q/AH 1Q/S ... ... 2Q/S ... ... ... 3E/F 4T/AH 1QQ/AH 3U/K 4Q/S 1QQ/AH ... ... 2E/F ... ... ... 4Q/K 4E/F 1QE/AH 4Q/F 4QT/AH 1QE/AH ... ...

3 ... ... ... 4Q/S 5E/AH 2 4T/K 5T/AH 2 ... ...

3Q/F ... ... ... 4U/K 5T/K 2E/AH ... ... ... ... ... 3Q/S ... ... ... 5Q/F 6Q/AH 2T/AH ... ... ... ... ... 3E/F ... ... ... 5T/K 6Q/S 2Q/S ... ... ... ... ...

4 ... ... ... 6 6QT/AH 2QQ/AH ... ... ... ... ...

Source: AISC Manual of Steel Construction, Allowable Stress Design, Ninth Edition, 1989.



STANDARD BOLTS, NUTS AND WASHERS Design Aid 11.3.1 Dimensions of Nuts and Bolts (Cont.)

Nuts

Nut dimensions, rounded to nearest Q/AH inch, are in accordance with ANSI B18.2.2 – 1972

Dimensions for Nuts

Square Hex Heavy Square Heavy Hex Nut Size Width

F Width

C Height

H Width

F Width

C Height

H Width

F Width

C Height

H Width

F Width

C Height

H in. in. in. in. in. in. in. in. in. in. in. in. in. Q/F U/AH T/K Q/F U/AH Q/S Q/F Q/S QQ/AH Q/F Q/S O/AH Q/F E/K T/K U/K T/AH O/AH T/K T/AH QQ/AH 1 E/K QQ/AH QE/AH E/K Q/S qE/AH 1Q/K U/AH E/F U/K U/AH U/K 1Q/F Q/S U/K 1 Q/S T/K 1 1U/AH O/AH QT/AH 1Q/AH O/AH 1Q/AH 1Q/S T/K 1Q/AH 1Q/F T/K E/F 1Q/K 1O/AH QQ/AH 1Q/K 1T/AH T/K 1Q/F 1E/F E/F 1Q/F 1U/AH E/F U/K 1T/AH 1U/K E/F 1T/AH 1Q/S E/F 1U/AH 2Q/AH U/K 1U/AH 1QQ/AH U/K 1 1Q/S 2Q/K U/K 1Q/S 1E/F U/K 1T/K 2T/AH 1 1T/K 1U/K 1

1Q/K 1QQ/AH 2E/K 1 1QQ/AH 1QT/AH 1 1QE/AH 2O/AH 1Q/K 1QE/AH 2Q/AH 1Q/K 1Q/F 1U/K 2T/K 1Q/K 1U/K 2E/AH 1Q/AH 2 2QE/AH 1Q/F 2 2T/AH 1Q/F 1E/K 2Q/AH 2QT/AH 1Q/F 2Q/AH 2E/K 1E/AH 2E/AH 3Q/K 1E/K 2E/AH 2Q/S 1E/K 1Q/S 2Q/F 3E/AH 1T/AH 2Q/F 2T/K 1T/AH 2E/K 3E/K 1Q/S 2E/K 2E/F 1Q/S 1E/F ... ... ... ... ... ... ... ... ... 2E/F 3E/AH 1E/F

2 ... ... ... ... ... ... ... ... ... 3Q/K 3T/K 2

2Q/F ... ... ... ... ... ... ... ... ... 3Q/S 4Q/AH 2E/AH 2Q/S ... ... ... ... ... ... ... ... ... 3U/K 4Q/S 2U/AH 2E/F ... ... ... ... ... ... ... ... ... 4Q/F 4QT/AH 2QQ/AH

3 ... ... ... ... ... ... ... ... ... 4T/K 5T/AH 2QT/AH

3Q/F ... ... ... ... ... ... ... ... ... 5 5E/F 3E/AH 3Q/S ... ... ... ... ... ... ... ... ... 5E/K 6E/AH 3U/AH 3E/F ... ... ... ... ... ... ... ... ... 5E/F 6T/K 3QQ/AH

4 ... ... ... ... ... ... ... ... ... 6Q/K 7Q/AH 3QT/AH




STANDARD BOLTS, NUTS AND WASHERS Design Aid 11.3.2 Dimensions of Standard Washers

Size of Bolt Size of Hole

Outside Diameter

Thickness Wire Gage

No.

Weight per 1,000 in Pounds

Size of Bolt Size of Hole

Outside Diameter

Thickness Wire Gage

No.

Plain Washers, U.S Standard (Dimensions in Inches)

Plain Washers, S.A.E. Standard (Dimensions in Inches)

Q/S O/AH 1E/K 12 (E/DS) 38.4 Q/S QU/DS 1Q/AH E/DS

O/AH T/K 1Q/S 12 (E/DS) 44.4 O/AH QO/DS 1E/AH E/DS

T/K QQ/AH 1E/F 10 (Q/K) 77.0 T/K WQ/DS 1T/AH E/DS

E/F QE/AH 2 10 (Q/K) 111.0 E/F QE/AH 1Q/S Q/K

U/K QT/AH 2Q/F 9 (T/DS) 153.0 U/K QT/AH 1E/F Q/K

1 1Q/AH 2Q/S 9 (T/DS) 176.0

1Q/K 1Q/F 2E/F 9 (T/DS)

1Q/F 1E/K 3 9 (T/DS)

Narrow – Gage Washers (Dimensions in Inches)

Q/S O/AH 1Q/F 12 (E/DS)

O/AH T/K 1E/K 12 (E/DS)

T/K QQ/AH 1Q/S 10 (Q/K)

E/F QE/AH 1E/F 10 (Q/K)

U/K QT/AH 2 9 (T/DS)

1 1Q/AH 2Q/F 9 (T/DS)

1Q/K 1Q/F 2Q/S 9 (T/DS)

1Q/F 1E/K 2E/F 9 (T/DS)

1E/K 1Q/S 3 8 (QQ/HF)




11.4 WELDING INFORMATION Design Aid 11.4.1 Weld Symbols Commonly Used in Precast Construction

Basic Welding Symbols and Their Meanings Elements of a Typical Weld Symbol

Location / Position of Symbol Supplemental Symbols

Type of Weld Arrow Side Other Side Both Sides Field Weld Weld All Around

Finishing Contours

Fillet Weld

Plug or Slot Weld

Not Applicable

Square

Side Designation

V

Bevel

Maximum Detailed Fillet Weld Sizes at Edges

Flare-Bevel

Groove Welds

Flare-V

End Returns

Stud Weld

No arrow or other side significance to the stud weld symbol.

For other basic and supplemental weld symbol and process information, refer to

ANSI/AWS A2.4 References a. AWS, Standard Symbols for Welding, Brazing and Nondestruction Examination, ANSI/AWS A2.4-86, American Welding Society, Miami,

Florida, 1986. b. AWS, Structural Welding Code – Steel, ANSI/AWS D1.1:2000, 17th Edition, American Welding Society, Miami, Florida, 2000. c. AISC, Load & Resistance Factor Design, – Manual of Steel Construction, Third Edition, American Institute of Steel Construction,

Chicago, Illinois, 2001.



WELDING INFORMATION Design Aid 11.4.2 Typical Welded Joints in Precast Construction

Fillet Weld

Fillet Weld

2 in. of T/AH fillet weld on 6 in. centers, each side

Q/F in. fillet weld, 6 in. long, each side

Plug Weld Flare Bevel with Fillet Weld

1 in. Ø plug welds x Q/S in. deep at 4 in. on-center Flare bevel weld of tube to plate followed by Q/F in. fillet weld reinforcing

Bevel with Fillet Weld Square Weld with Fillet

Q/F in. bevel weld with T/AH in. fillet weld reinforcing, one side

E/K in. square weld with Q/S in. fillet weld reinforcing, both sides

Stud Weld

Reinforcing Bar Welds

6 - Q/S in. Ø studs spaced at 3 in. on-center in one line



WELDING INFORMATION Design Aid 11.4.3 Properties of Weld Groups Treated as Lines (tw = 1)

Section b = width; d = depth

Distance to centroid

Section modulus xI / y

Polar moment of inertia IP, about center of gravity

(1) S =

2d6

Ip = 3d

12

(2)

S = 2d

3 Ip =

2 2d(3b d )6

+

(3)

S = bd Ip = 2 2b(3d b )6

+

(4)

2dy2(b d)

=+

2bx2(b d)

=+

Stop = 24bd d

6+

Sbott = 2d (4b d)6(2b d)

++

Ip =

4 2 2(b d) 6b d12(b d)

+ −+

(5)

2bx2b d

=+

S = 2dbd

6+

Ip =

3 2 3 48b 6bd d b12 2b d

+ + −+

(6)

2dyb 2d

=+

Stop =

22bd d3+

Sbott = 2d (2b d)3(b d)

++

Ip =

3 2 3 4b 6b d 8d d12 2d b

+ + −+

(7)

S = 2dbd

3+ Ip =

3(b d)6

+

(8)

2dyb 2d

=+

Stop =

22bd d3+

Sbott = 2d (2b d)3(b d)

++

Ip =

3 3 4b 8d d12 b 2d+ −

+

(9)

S = 2dbd

3+ Ip =

3 2 3b 3bd d6

+ +

(10)

S = 2rπ Ip = 32 rπ



11.5 SECTION PROPERTIES Design Aid 11.5.1 Properties of Geometric Sectionsa

SQUARE

Axis of Moments Through Center

A = d2

c = d2

I = 4d

12

S = 3d6

r = d 0.288675d12

=

RECTANGLE

Axis of Moments on Diagonal

A = bd

c = 2 2

bd

b d+

I = 3 3

2 2b d

6(b d )+

S = 2 2

2 2

b d

6 b d+

r = 2 2

bd

6(b d )+

SQUARE Axis of Moments on Base

A = d2 c = d

I = 4d

3

S = 3d

3

r = d 0.577350d3

=

RECTANGLE

Axis of Moments Any Line Through Center of Gravity

A = bd

c = bsina dcosa2+

I = 2 2 2 2bd(b sin a d cos a)

12+

S = 2 2 2 2bd(b sin a d cos a)

6(bsina dcosa)++

r = 2 2 2 2b sin a d cos a

12+

SQUARE Axis of Moments on Diagonal

A = d2

c = d 0.707107d2

=

I = 4d

12

S = 3

3d 0.117851d6 2

=

r = d 0.288675d12

=

HOLLOW RECTANGLE


A = bd – b1d1

c = d2

I = 3 3

1 1bd b d12−

S = 3 3

1 1bd b d6d−

r = 3 3

1 1bd b d12A

−

RECTANGLE Axis of Moments Through

Center

A = bd

c = d2

I = 3bd

12

S = 2bd

6

r = d 0.288675d12

=

EQUAL RECTANGLES

Axis of Moments Through Center of Gravity

A = b(d – d1)

c = d2

I = 3 3

1b(d d )12

−

S = 3 3

1b(d d )6d

−

r = 3 3

1

1

d d12(d d )

−−

a. Source: “Manual of Steel Construction, Allowable Stress Design,” Ninth Edition, American Institute of Steel Construction, Chicago, IL, 1989.



SECTION PROPERTIES Design Aid 11.5.1 Properties of Geometric Sections (Cont.)

RECTANGLE

Axis of Moments on Base

A = bd

c = d2

I = 3bd

3

S = 2bd

3

r = d 0.577350d3

=

UNEQUAL RECTANGLES


A = bt + b1t1

c = 2

1 1 1bt b t (d t )A

+ −Q/S Q/S

I = 33

2 21 11 1 1

b tbt bty b t y12 12

+ + +

S = 11

I ISc c

=

r = IA

TRIANGLE Axis of Moments Through

Center of Gravity

A = bd2

c = 2d3

I = 3bd

36

S = 2bd

24

r = d18

HALF CIRCLE

Axis Through Moments of Center of Gravity

A = 2R

2π

c = 4R 13

⎛ ⎞−⎜ ⎟π⎝ ⎠

I = 4 8R8 9π⎛ ⎞−⎜ ⎟π⎝ ⎠

S = 3 2R (9 64)

24 (3 4)⎡ ⎤ ⎡ ⎤π −⎢ ⎥ ⎢ ⎥

π −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

r = 29 64R6

π −π

TRIANGLE

Axis of Moments on Base

A = bd2

c = d

I = 3bd

12

S = 2bd

12

r = d6

PARTIAL CIRCLE

Axis of Moments Through Circle Center

NOTE: Angles in Radians

I = 4

2 2 311

yR (R y )8 2

π + −

22 2 2 1 1

1 1yR y R y R sin

4 R−⎛ ⎞− − +⎜ ⎟

⎝ ⎠

A = 2

2 21 1

R y R y2

π − −

2 1 1yR sin

R− ⎛ ⎞− ⎜ ⎟⎝ ⎠

c = 2 2 3 / 2

12(R y )/ A

3−

TRAPEZOID


A = 1d(b b )2+

c = 1

1

d(2b b )3(b b )

++

I = 3 2 2

1 1

1

d (b 4bb b )36(b b )

+ ++

S = 2 2 2

1 1

1

d (b 4bb b )12(2b b )

+ ++

r = 2 2

1 11

d x 2(b 4bb b )6(b b )

+ ++

PARABOLA

A = 4 ab

3

m = 2 a5

I1 = 316 a b175

I2 = 34 ab15

I3 = 332 a b105




CIRCLE


A = 2

2d R4

π = π

c = d R2

=

I = 4 4d R

64 4π π=

S = 3 3d R

32 4π π=

r = d R4 2

=

HALF PARABOLA

A = 2 ab3

m = 2 a5

n = 3 b8

I1 = 38 a b175

I2 = 319 ab480

I3 = 316 a b105

I4 = 32 ab15

HOLLOW CIRCLE


A = 2 2

1(d d )4

π −

c = d2

I = 4 4

1(d d )64

π −

S = 4 4

1(d d )32d

π −

r = 2 2

1d d4−

COMPLEMENT OF HALF

PARABOLA

A = 1 ab3

m = 7 a10

n = 3 b4

I1 = 337 a b2100

I2 = 31 ab80

PARABOLIC FILLET

IN RIGHT ANGLE

a = t2 2

b = t2

A = 21 t6

m = n = 4 t5

I1 = I2 = 411 t2100

*ELLIPTIC COMPLEMENT

* See note on next page

A = ab 14π⎛ ⎞−⎜ ⎟

⎝ ⎠

m = a

6 14π⎛ ⎞−⎜ ⎟

⎝ ⎠

n = b

6 14π⎛ ⎞−⎜ ⎟

⎝ ⎠

I1 = 3 1 1a b3 16 36 1

4

⎛ ⎞⎜ ⎟π⎜ ⎟− −⎜ ⎟π⎛ ⎞−⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

I2 = 3 1 1ab3 16 36 1

4

⎛ ⎞⎜ ⎟π⎜ ⎟− −⎜ ⎟π⎛ ⎞−⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠




*HALF ELLIPSE

A = 1 ab2

π

m = 4a3π

I1 = 3 8a b8 9π⎛ ⎞−⎜ ⎟π⎝ ⎠

I2 = 31 ab8

π

I3 = 31 a b8

π

REGULAR POLYGON

n = number of sides

φ = 180n

a = 2 212 R R−

R = a2sinφ

R1 = a2 tanφ

A = 21 na cot4

φ

= 2 21

1 nR sin2 nR tan2

φ = φ

I1 = I2 =2 2A(6R a )

24−

= 2 21A(12R a )

48+

r1 = r2 =2 26R a24

−

= 2

112R a48

+

*QUARTER ELLIPSE

A = 1 ab4

π

m = 4a3π

n = 4b3π

I1 = 3 4a b16 9π⎛ ⎞−⎜ ⎟π⎝ ⎠

I2 = 3 4ab16 9π⎛ ⎞−⎜ ⎟π⎝ ⎠

I3 = 31 a b16

π

I4 = 31 ab16

π

BEAMS AND CHANNELS I3 = 2 2

x yI sin I cosφ + φ

I4 = 2 2x yI cos I sinφ + φ

fb = x y

y xM sin cosI I

⎛ ⎞⎜ φ + φ⎟⎜ ⎟⎝ ⎠

Where M is bending moment due to force F.

* To obtain properties of half circles, quarter circle and circular complement, substitute a = b = R.




ANGLE

AXIS OF MOMENTS THROUGH CENTER OF GRAVITY

Z-Z is Axis of Minimum I

tan2θ = y x

2kI I−

A = t(b – c) x = 2b ct

2(b c)++

y = 2d at

2(b c)++

K = product of inertia about X-X & Y-Y

= abcdt4(b c)

±+

Ix = 3 3 31 [t(d y) by a(y t) ]3

− + − −

Iy = 3 3 31 [t(b x) dx c(x t) ]3

− + − −

Iz = 2 2x yI sin I cos K sin2θ + θ + θ

Iw = 2 2x yI cos I sin K sin2θ + θ + θ

K is negative when heel of angle, with respect to center of gravity, is in first or third quadrant, positive when in second or fourth quadrant.



SECTION PROPERTIES Design Aid 11.5.2 Plastic Section Moduli and Shape Factors

SECTION PLASTIC MODULUS, Z3 , in.3 SHAPE FACTOR

2bh4

1.5

x-x axis

2wbt(h t) (h 2t)4

− + −

1.12 (approx.)

y-y axis 2 2b t (h 2t)w2 4

−+

1.55 (approx.)

2w(h 2t)bt(h t)4−− + 1.12 (approx.)

3h6

1.70

33h 2t1 16 h

⎡ ⎤⎛ ⎞⎢ ⎥− −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

th2 for t << h

3

4

2t1 116 h3 2t1 1

h

⎡ ⎤⎛ ⎞⎢ ⎥− −⎜ ⎟⎝ ⎠⎢ ⎥

⎢ ⎥π ⎛ ⎞⎢ ⎥− −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

1.27 for t << h

22bh 2w 2t1 1 14 b h

⎡ ⎤⎛ ⎞⎛ ⎞⎢ ⎥− − −⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦ 1.12 (approx.) for thin walls

2bh12

2

For TS shapes, refer to AISC-LRFD Manual, Third Edition for Zs values.



11.6 METRIC CONVERSION Design Aid 11.6.1 Metric Calculations and Example

“Hard” SI (metric) Calculations for Precast Concrete

Design Aid 11.4.1 shows direct conversion from inch-lb. (U.S. customary) to SI units. Calculations made in SI units are usually rounded to even numbers. These are known as “hard” metric calculations. The metric version of the Code is 318M-02. Some of the common SI units used are: Concrete Strength: 20 MPa is approximately equivalent to 3000 psi 35 MPa is approximately equivalent to 5000 psi Concrete weight (or mass): Normal weight concrete may be assumed to weigh 2400 kg/m3 (149.8 pcf). Lightweight concrete may be assumed to weigh 1900 kg/m3 (118.6 pcf). Reinforcement: Most U. S. reinforcing bar manufacturers mark bars with metric designations, although the actual sizes have not changed (see Design Aid 11.2.7). Grade 420 reinforcing bar (fy = 420 MPa) is equivalent to Grade 60 reinforcing bar. Prestressing Strand: Strand diameters and areas are rounded to the nearest mm, e.g., 13 mm is equivalent to Q/S in. diam-eter, area = 99 mm2. Relationships in SI

Structural engineering calculations in SI units involve forces which include gravitational effects, rather than just weights, or mass. Thus one kilogram (kg) of mass converts to 9.8 newtons (N) of force. For example, a 50 mm thick concrete topping 1 meter wide weighs (50/1000)m × 1m × 2400 kg/m3 = 120 kg/m. It exerts 120 × 9.8 = 1176 N/m or 1.18 kN/ m of force.

Pressure or stress is expressed in pascals (P). 1P = 1N/m2. It is more common to work in megapascals (MPa). 1 MPa =1N/mm2. Bending moments are expressed in newton-meters (N-m) or kilonewton-meters (kN-m).

Design Aid 11.6.2 lists quantities that are fre-quently encountered in the design of precast concrete as well as in general structural engineering practice. Most U.S. Government agencies that require SI unit dimensioning of contract documents will also require consistent use of the listed SI units given in the table. The equivalent U.S. customary units listed are those traditionally used by the design professions in the U.S.

Conversion of frequently encountered concrete stress coefficients used in ACI 318-02 and the PCI Handbook are tabulated in Design Aid 11.6.4.

Example Use of Eq. 18-3 from 318M-02 (Similar to Example 4.2.2.3) Given: Double tee similar to PCI standard 8DT24+2 Concrete: Precast cf ′ = 35 MPa Topping cf ′ = 20 MPa Reinforcement: 12-13 mm dia. 1860 MPa low relaxation strands (6 ea. stem). Area per strand = 99 mm2 Aps = 12(99) = 1188 mm2 As = 2 – #19 = 568 mm2 fy = 420 MPa Problem: Find the design flexural strength of the composite section using Eq. 18-3 from ACI 318M-02.

fps = p pupu p

1 p p

f df 1 ( )f d

⎛ ⎞⎡ ⎤γ⎜ ⎟′− ρ + ω − ω⎢ ⎥⎜ ⎟′β ⎢ ⎥⎣ ⎦⎝ ⎠

′ω = 0 in this example

ρp = ps

p

A 1188 0.00087bd 2400(570)

+ =

ω = ys

c

fA 568 420 0.0080bd f 2400(620) 20

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟′ ⎝ ⎠⎝ ⎠

fps = 0.28 1860 6201860 1 0.00087 (0.0080)0.85 20 570

⎛ ⎞⎡ ⎤− +⎜ ⎟⎢ ⎥⎣ ⎦⎝ ⎠

= 1805 MPa

a = ps ps s y

c

A f A f 1188(1805) 568(420)0.85f b 0.85(20)2400

+ −=′

a = 47 mm

Mn = ps ps p s ya aA f d A f d2 2

⎛ ⎞ ⎛ ⎞− + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=

570 24 620 241188(1805) 568(420)1000 1000

− −⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 2 2 2 22x(d x) d 2(c e ) [e x c (d x)]d

⎧ ⎫− + − + − + −⎨ ⎬⎩ ⎭

2 3(x d) (4e d x) d ]− − + − − = 1,312,991 N-m = 1313 kN-m φMn = 0.9(1313) = 1182 kN-m



METRIC CONVERSION Design Aid 11.6.2 Conversion from U.S. Customary Units to International System (SI)

To convert from

to multiply by

Length inch (in.) millimeter (mm) 25.4 inch (in.) meter (m) 0.0254 foot (ft) meter (m) 0.3048 yard (yd) meter (m) 0.9144 mile (mi)

kilometer (km) 1.6093

Area square foot (ft2) square meter (m2) 0.09290 square inch (in.2) square millimeter (mm2) 645.2 square inch (in.2) square meter (m2) 0.0006452 square yard (yd2) square meter (m2) 0.8361 acre (A) hectare (ha) = 10,000 m2 0.4047 square mile

square kilometer 2.590

Volume cubic inch (in.3) cubic meter (m3) 0.00001639 cubic foot (ft3) cubic meter (m3) 0.02832 cubic yard (yd3) cubic meter (m3) 0.7646 gallon (gal) Can. liquida liter 4.546 gallon (gal) Can. liquida cubic meter (m3) 0.004546 gallon (gal) U.S. liquida liter 3.785 gallon (gal) U.S. liquida

cubic meter (m3) 0.003785

Force kip kilogram (kgf) 453.6 kip newton (N) 4448.0 pound (lb) kilogram (kgf) 0.4536 pound (lb)

newton (N) 4.448

Pressure or Stress kips/square inch (ksi) megapascal (MPa)b 6.895 pound/square foot (psf) kilopascal (kPa)b 0.04788 pound/square inch (psi) kilopascal (kPa)b 6.895 pound/square inch (psi) megapascal (MPa)b 0.006895 pound/square foot (psf)

kilogram/square meter (kgf/m2) 4.882

Mass pound (avdp) kilogram (kg) 0.4536 ton (short, 2000 lb) kilogram (kg) 907.2 ton (short, 2000 lb) tonne (t) 0.9072 grain kilogram (kg) 0.00006480 tonne (t) kilogram (kg) 1000

a. One U.S. gallon equals 0.8321 Canadian gallon. b. A pascal equals one newton/square meter; 1 Pa = 1 N/m2. Note: To convert from SI units to U.S. customary units (except for temperature), divide by the factors given in this table.



METRIC CONVERSION Design Aid 11.6.2 Conversion from U.S. Customary Units to International System (SI) (Cont.)

To convert from

to multiply by

Mass (weight) per Length kip/linear foot (klf) kilogram/meter (kg/m) 1488 pound/linear foot (plf) kilogram/meter (kg/m) 1.488 pound/linear foot (plf)

newton/meter (N/m) 14.593

Mass per volume (density) pound/cubic foot (pcf) kilogram/cubic meter (kg/m3) 16.02 pound/cubic yard (pcy)

kilogram/cubic meter (kg/m3) 0.5933

Bending Moment or Torque pound-inch (lb-in.) newton-meter 0.1130 pound-foot (lb-ft) newton-meter 1.356 kip-foot (kip-ft)

newton-meter 1356

Temperature degree Fahrenheit (°F) degree Celsius (°C) tC = (tF – 32)/1.8 degree Fahrenheit (°F)

degree Kelvin (K) tK = (tF + 459.7)/1.8

Energy British thermal unit (Btu) joule (j) 1056 kilowatt-hour (kwh)

joule (j) 3,600,000

Power horsepower (hp) (550 ft lb/sec.)

watt (W) 745.7

Velocity mile/hour (mph) kilometer/hour 1.609 mile/hour (mph)

meter/second (m/s) 0.4470

Other Section modulus (in.3) mm3 16,387 Moment of inertia (in.4) mm4 416,231 Coefficient of heat transfer (Btu/ft2/h/°F) W/m2/°C 5.678 Modulus of elasticity (psi) MPa 0.006895 Thermal conductivity (Btu/in./ft2/h/°F) Wm/m2/°C 0.1442 Thermal expansion in./in./°F mm/m2/°C 1.800 Area/length (in.2/ft) mm2/m 2116.80



METRIC CONVERSION Design Aid 11.6.3 Preferred SI Units and U.S. Customary Equivalents

Quantity SI U.S. customary

Area, cross section mm2 in.2 Area, plan dimension mm2 ft2 Bending moment kN-m kip-ft Coefficient of thermal expansion mm/(mm-°C) in./in./°F Deflection mm in. Density, linear kg/m lb/ft, kip/ft Density, area kg/m2 lb/ft2, kip/ft2 Density, mass kg/m3 lb/ft3, kip/ft3 Force kN lb, kip Force, per unit length kN/m lb/ft, kip/ft Force, per unit area kN/m2 lb/ft2, kip/ft2 Length, cross section mm in. Length, plan dimension mm ft Mass kg lb, kip Modulus of elasticity MPa psi, ksi Moment of inertia 106 mm4 in.4 Section modulus 106 mm3 in.3 Stress MPa psi, ksi Temperature °C °F Torque kN-m lb-ft, kip-ft

Design Aid 11.6.4 Concrete stress coefficents

U.S. customary coefficient SI coefficient U.S. customary

coefficient SI coefficient

0.5 0.04 3.3 0.27 0.6 0.05 3.5 0.29

0.667 0.06 4.0 0.33 1.0 0.08 4.4 0.37 1.1 0.09 5.0 0.42 1.2 0.10 5.5 0.46 1.25 0.10 6.0 0.50 1.5 0.12 6.3 0.52 1.6 0.13 6.5 0.54 1.7 0.14 7.0 0.58 1.9 0.16 7.5 0.62 2.0 0.17 8.0 0.66 2.4 0.20 10.0 0.83 3.0 0.25 12.0 1.00

Examples: U.S. customary: psi SI equivalent: 0.08 MPa U.S. customary: 10 psi SI equivalent: 0.83 MPa


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